Corrected exercises: infinite series, 201-SN3-RE (201-NYB)

The series $\sum a_{n}$ converges to $S$ if the SEQUENCE $S_{n}=a_{1}+\cdots+a_{n}$ tends to $S$: the convergence of a series is, by definition, the convergence of a sequence, the one you see flattening against the dashed line in the figure. For the geometric series: $S_{n}=a+ar+\cdots+ar^{n-1}$ and $rS_{n}=ar+\cdots+ar^{n}$; subtracting, everything telescopes except the ends: $S_{n}(1-r)=a(1-r^{n})$, hence $S_{n}=a\,\frac{1-r^{n}}{1-r}$ (for $r\ne 1$).

Everything rests on $r^{n}$: if $|r|<1$, $r^{n}\to 0$ and $S_{n}\to\frac{a}{1-r}$; if $|r|\ge 1$, $r^{n}$ does not tend to 0 and the series diverges. Here $a=5$, $r=\frac{2}{3}$: $S=\frac{5}{1-\frac{2}{3}}=15$, the line in the figure. Starting at $n=2$: the first term summed is $5\left(\frac{2}{3}\right)^{2}=\frac{20}{9}$, so $S=\frac{20/9}{1/3}=\frac{20}{3}\approx 6.67$ (or else $15-5-\frac{10}{3}$, same result). The formula $\frac{a}{1-r}$ requires knowing WHICH term comes first.

$0.999\ldots=\frac{9}{10}+\frac{9}{100}+\cdots=\sum_{n=1}^{\infty}\frac{9}{10}\left(\frac{1}{10}\right)^{n-1}=\frac{9/10}{1-1/10}=1$: exactly 1, not “almost”: the two notations name the same number. Likewise $0.727272\ldots=\frac{72}{100}+\frac{72}{100^{2}}+\cdots=\frac{72/100}{1-1/100}=\frac{72}{99}=\frac{8}{11}$. Every repeating decimal is a geometric series in disguise, hence a rational number.

The ball falls 2 m, then for each bounce of height $h$ it goes up AND comes back down: $D=2+2\sum_{n=1}^{\infty}2(0.6)^{n}=2+4\cdot\frac{0.6}{1-0.6}=2+6=8$ m. Infinitely many bounces, eight metres: the heights decrease geometrically, and a geometric series with $|r|<1$ has a finite sum. This is the modern answer to Zeno's paradox: infinitely many steps do not imply an infinite distance (or time), as soon as the steps shrink fast enough.

If $a_{n}\not\to 0$, the series diverges: the partial sums keep receiving additions that never die out. Here $\frac{n}{2n+1}\to\frac{1}{2}\ne 0$: DIVERGENT, in one line. What the test does not say: if $a_{n}\to 0$, you can conclude NOTHING: it is not a convergence test, only a filter for gross divergence. The harmonic series of c) is the official counterexample to the converse.

$\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$. The partial sum collapses like dominoes: $S_{n}=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots+\left(\frac{1}{n}-\frac{1}{n+1}\right)=1-\frac{1}{n+1}$, each negative term cancelling the next positive one. Limit: $S=1$. The telescoping series is, with the geometric series, the only one whose EXACT sum you obtain by computing the partial sums directly.

Group the terms: $\frac{1}{3}+\frac{1}{4}>\frac{2}{4}=\frac{1}{2}$; then $\frac{1}{5}+\cdots+\frac{1}{8}>\frac{4}{8}=\frac{1}{2}$; the next block of 8 terms again exceeds $\frac{1}{2}$, and so on: each block of $2^{k}$ terms, bounded below by its last term, contributes more than one half. With infinitely many blocks, $S_{n}$ exceeds any number: DIVERGENCE. The term $\frac{1}{n}$ does tend to 0, but not fast enough: exactly the divergent $\int\frac{dx}{x}$ of improper integrals, whose portrait is the curve in the figure.

$H_{n}>10$ requires $\ln n>9.42$, that is $n>e^{9.42}\approx 12\,400$ terms. To exceed 20: $n>e^{19.42}\approx 2.7\times 10^{8}$, a quarter of a billion terms to gain ten more units. The harmonic series diverges, but at a spectacularly slow, logarithmic pace. No calculator adding the terms one by one will ever “see” this divergence: only the proof guarantees it, and that is the whole difference between computing and proving.

Draw the rectangles of width 1 and height $a_{n}$: placed to the left of the integers, they stick out above the curve and $\sum a_{n}\ge\int_{1}^{\infty}f$; placed to the right, they stay below it and $\sum_{n\ge 2}a_{n}\le\int_{1}^{\infty}f$. The series is therefore squeezed between the integral and the integral plus the first term: they converge or diverge TOGETHER. For $\frac{1}{n^{2}}$: $\int_{1}^{\infty}\frac{dx}{x^{2}}=1$ converges (improper integrals chapter), so $\sum\frac{1}{n^{2}}$ CONVERGES, and its sum lies between 1 and 2.

The test carries over, word for word, the $p$ criterion for integrals: $\int_{1}^{\infty}\frac{dx}{x^{p}}$ converges if and only if $p>1$, so $\sum\frac{1}{n^{p}}$ does too. $\sum\frac{1}{\sqrt{n}}$: $p=\frac{1}{2}$, diverges. $\sum\frac{1}{n^{1.01}}$: $p=1.01>1$, converges: the boundary is exactly at $p=1$, and a hair on the right side is enough. $\sum\frac{1}{n^{3}}$: converges. The harmonic series is the boundary case $p=1$, on the losing side.

Remainder after 100 terms: $R_{100}\le\int_{100}^{\infty}\frac{dx}{x^{2}}=\frac{1}{100}=0.01$ (and $R_{100}\ge\frac{1}{101}$). Stopping at 100 terms therefore guarantees an error of at most $0.01$, which is not enough to guarantee two correct decimals. Consistency: $\frac{\pi^{2}}{6}\approx 1.6449$, and the sum of the first 100 terms is about $1.6350$: the gap $0.0099$ falls right inside the bounds $[\frac{1}{101},\frac{1}{100}]$. Bounding the remainder turns a qualitative test into a tool for controlled numerical computation.

$\frac{1}{n^{2}+n}<\frac{1}{n^{2}}$, a convergent series bounds it from above: CONVERGES (it is in fact the telescoping series of exercise 2, with sum exactly 1: the comparison is confirmed by the direct computation). $\frac{\ln n}{n}\ge\frac{1}{n}$ as soon as $n\ge 3$ (since $\ln n\ge 1$): bounded below by the divergent harmonic series, it DIVERGES. A logarithm in the numerator makes a divergence worse; in the denominator, it would not be enough to rescue $\frac{1}{n\ln n}$ either: only a true power $p>1$ rescues a series.

Test: let $a_{n}>0$ and $b_{n}>0$. If $\lim_{n\to\infty}\frac{a_{n}}{b_{n}}=c$ with $0<c<\infty$, then $\sum a_{n}$ and $\sum b_{n}$ behave the SAME WAY: both converge or both diverge. The reason in one sentence: if the ratio tends to $c>0$, then from some index on $a_{n}$ is squeezed between $\frac{c}{2}b_{n}$ and $2c\,b_{n}$, that is, bounded by two multiples of $b_{n}$; multiplying a series by a nonzero constant never changes whether it converges, and the squeeze passes on convergence in one direction and divergence in the other. The limiting cases deserve a mention: if $c=0$, the convergence of $\sum b_{n}$ implies that of $\sum a_{n}$ but nothing more; if $c=\infty$, the divergence of $\sum b_{n}$ implies that of $\sum a_{n}$. Only $0<c<\infty$ gives the full equivalence.

First series: at infinity, the numerator behaves like $2n^{2}$ and the denominator like $n^{4}$, so the term behaves like $\frac{2}{n^{2}}$. Take $b_{n}=\frac{1}{n^{2}}$: $\frac{a_{n}}{b_{n}}=\frac{n^{2}(2n^{2}+3)}{n^{4}-n+5}\to 2$, a strictly positive, finite constant. Since $\sum\frac{1}{n^{2}}$ converges (p-series, $p=2>1$), the series CONVERGES. Numerical check: the ratio equals $2.0000000$ for $n=10^{6}$. Second series: $\sqrt{n^{2}+1}$ behaves like $n$, so take $b_{n}=\frac{1}{n}$: $\frac{a_{n}}{b_{n}}=\frac{n}{\sqrt{n^{2}+1}}\to 1$. Since the harmonic series DIVERGES, the series DIVERGES. The reference is always chosen by the same rule: keep only the dominant terms, top and bottom.

Direct comparison: you would like to write $\frac{1}{n^{2}-n}\le\frac{1}{n^{2}}$ and conclude by bounding from above. But that is FALSE: since $n^{2}-n<n^{2}$, the quotient is LARGER, not smaller. Numerically, at $n=5$: $\frac{1}{20}=0.05$ against $\frac{1}{25}=0.04$. The inequality falls on the wrong side, and bounding by a convergent series is impossible in that direction. Limit comparison: $\frac{1/(n^{2}-n)}{1/n^{2}}=\frac{n^{2}}{n^{2}-n}=\frac{1}{1-\frac{1}{n}}\to 1$, a strictly positive constant. So the series behaves like $\sum\frac{1}{n^{2}}$: it CONVERGES. The failure of direct comparison is instructive because it reveals what that test really requires: an inequality that is TRUE FOR EVERY INDEX and points in the right direction, a rigid condition that the slightest change in the denominator breaks. The limit comparison test, on the other hand, looks only at the ASYMPTOTIC behaviour and ignores both the first indices and the direction of the inequality. That is why it is almost always the first tool to reach for in front of a quotient of polynomials or roots.

The equivalent terms come from Maclaurin series, or equivalently from the classic limits of the L'Hôpital chapter. First series: $\sin u\approx u$ as $u\to 0$, so with $u=\frac{1}{n}$, compare with $b_{n}=\frac{1}{n}$: $\frac{\sin(1/n)}{1/n}\to 1$, exactly the fundamental limit $\frac{\sin x}{x}\to 1$ read at $x=\frac{1}{n}$. Since the harmonic series diverges, $\sum\sin\left(\frac{1}{n}\right)$ DIVERGES. Check: the sum of the first $10^{5}$ terms is about $11.90$, to compare with $\ln(10^{5})\approx 11.51$: the characteristic logarithmic growth is indeed there. Second series: $1-\cos u\approx\frac{u^{2}}{2}$, so compare with $b_{n}=\frac{1}{2n^{2}}$: the ratio tends to 1. Since $\sum\frac{1}{n^{2}}$ converges, $\sum\left(1-\cos\left(\frac{1}{n}\right)\right)$ CONVERGES. Check: the sum of the first $10^{5}$ terms is about $0.7788$, a value that stays below $\sum\frac{1}{2n^{2}}=\frac{\pi^{2}}{12}\approx 0.8225$, as it should since $1-\cos u$ is strictly less than $\frac{u^{2}}{2}$. The message of the chapter is complete here: Taylor expansions BUILD the reference series, and the tests only have to conclude. Two expressions as close as $\sin\left(\frac{1}{n}\right)$ and $1-\cos\left(\frac{1}{n}\right)$ fall on either side of the boundary, because one dies out at order 1 and the other at order 2.

Linearity: if $\sum a_{n}$ and $\sum b_{n}$ CONVERGE, with sums $A$ and $B$, then for all real $\lambda$ and $\mu$ the series $\sum(\lambda a_{n}+\mu b_{n})$ converges and equals $\lambda A+\mu B$. The proof is immediate on the partial sums, which are finite sums where linearity holds, and the limit of a sum is the sum of the limits. Computation: $\sum_{n=1}^{\infty}\frac{1}{2^{n}}$ is geometric with first term $\frac{1}{2}$ and ratio $\frac{1}{2}$, so it equals $\frac{1/2}{1-1/2}=1$; and $\sum_{n=1}^{\infty}\frac{1}{3^{n}}=\frac{1/3}{1-1/3}=\frac{1}{2}$. By linearity, the requested sum is $1+\frac{1}{2}=\frac{3}{2}$, which the numerical computation confirms.

The first equality is correct: all the terms equal 0, the series converges and its sum is 0. The second is ILLEGITIMATE, because it splits a convergent series into the DIFFERENCE of two divergent series, which linearity does not allow: that property assumes both series converge BEFORE they are combined. Writing $\sum\frac{1}{n}-\sum\frac{1}{n}$ amounts to writing $\infty-\infty$, an indeterminate form with no value. The exact rule has three cases. If $\sum a_{n}$ and $\sum b_{n}$ converge, $\sum(a_{n}+b_{n})$ converges, to the sum of the sums. If one converges and the other diverges, $\sum(a_{n}+b_{n})$ necessarily DIVERGES, because if it converged, subtracting the convergent one would make the divergent one convergent, a contradiction. Finally, if both diverge, NOTHING can be said: the example in the statement gives a convergent sum, whereas $a_{n}=b_{n}=\frac{1}{n}$ gives $\sum\frac{2}{n}$, divergent. It is the only undecidable case, and it is the one the student ran into.

Partial sums: $S_{1}=1$, $S_{2}=0$, $S_{3}=1$, $S_{4}=0$, and so on, alternating forever. The sequence $(S_{n})$ has no limit, it oscillates between 0 and 1: the series DIVERGES. You can also conclude in one line with the divergence test, since $(-1)^{n}$ does not tend to 0. Groupings: pairing the terms two by two from the first one, $(1-1)+(1-1)+(1-1)+\cdots=0+0+0+\cdots=0$. Leaving the first term on its own, $1-(1-1)-(1-1)-\cdots=1-0-0-\cdots=1$. The two computations use exactly the same terms in the same ORDER, and yet they give 0 and 1. The rule this brings out is that GROUPING with parentheses, that is, associativity, is NOT valid for a divergent series. The correct theorem goes one way only: if a series converges, every grouping of its terms converges to the same sum; but a convergent grouping never proves that the original series converges. In other words, parentheses can CREATE convergence where there is none, and that is exactly what the first grouping above does. Historically, this series occupied mathematicians for more than a century, and it is what forced the modern definition through partial sums.

Let $\sum a_{n}$ be a series and $N$ a fixed index. The partial sums beyond $N$ can be written $S_{n}=S_{N}+\left(a_{N+1}+\cdots+a_{n}\right)$, where $S_{N}$ is a CONSTANT. The sequence $(S_{n})$ therefore converges if and only if the second part converges, that is, if and only if the series $\sum_{n>N}a_{n}$ converges. Changing, removing or adding a finite number of terms only shifts that constant: WHETHER the series converges is unchanged, the SUM is shifted by the same amount. Illustration: $\sum_{n=1}^{\infty}\frac{1}{2^{n}}=1$; removing the first three terms leaves $1-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\right)=1-\frac{7}{8}=\frac{1}{8}$. Still convergent, different sum. Remainder criterion: write $R_{N}=\sum_{n>N}a_{n}$ for the remainder of order $N$. A series converges if and only if its remainders are defined and $R_{N}\to 0$ as $N\to\infty$. Here $R_{N}=\frac{1}{2^{N}}\to 0$ ✓. This criterion has a practical consequence used all the time: to study whether a series converges, you may ignore the first terms without a second thought, which allows comparisons that hold only “from some index on”, as in part c) of the previous exercise.

$\frac{a_{n+1}}{a_{n}}=\frac{n+1}{2^{n+1}}\cdot\frac{2^{n}}{n}=\frac{1}{2}\cdot\frac{n+1}{n}\to\frac{1}{2}<1$: CONVERGENT. Intuition: the factor $n$ weighs nothing against $2^{n}$: at infinity, the series behaves like a geometric series with ratio $\frac{1}{2}$, barely inflated. (Its exact sum is 2, by the way, a classic of power series.) The ratio test measures precisely “which geometric series the series resembles”.

$\frac{2^{n+1}}{(n+1)!}\cdot\frac{n!}{2^{n}}=\frac{2}{n+1}\to 0<1$: overwhelming convergence (the sum is $e^{2}\approx 7.39$, the exponential series). For $\frac{n!}{10^{n}}$: $\frac{(n+1)!}{10^{n+1}}\cdot\frac{10^{n}}{n!}=\frac{n+1}{10}\to\infty$: DIVERGENT, the general term even blows up. Verdict of the duel: the factorial ALWAYS ends up crushing any exponential $c^{n}$: its factors grow ($n+1$), those of the exponential are fixed ($c$). From $n=10$ on, each new factor of $n!$ exceeds 10 and the race is over.

For $\frac{1}{n}$: $\frac{n}{n+1}\to 1$. For $\frac{1}{n^{2}}$: $\left(\frac{n}{n+1}\right)^{2}\to 1$. Same verdict $L=1$, but the first DIVERGES and the second CONVERGES: the case $L=1$ allows NO conclusion, and unfortunately it covers every p-series, the heart of the chapter. The tools that take over: the integral test and the comparison tests, which tell powers apart finely where the ratio is blind. Practical rule: the ratio test for factorials and $c^{n}$; p-series and comparison for powers of $n$.

$\frac{a_{n+1}}{a_{n}}=\frac{(n+1)^{n+1}}{(n+1)!}\cdot\frac{n!}{n^{n}}=\frac{(n+1)^{n}}{n^{n}}=\left(1+\frac{1}{n}\right)^{n}\to e\approx 2.72>1$: DIVERGENT. Along the way, the result says that $n^{n}$ grows faster than $n!$ by a factor of $e$ per step: the seed of Stirling's formula. A question that tests both the test and the fundamental limit defining the number $e$: exactly the exam format.

If $b_{n}$ is DECREASING and tends to 0, then $\sum(-1)^{n+1}b_{n}$ converges. Here $b_{n}=\frac{1}{n}$: decreasing, limit zero: convergence guaranteed (to $\ln 2$, a value taken as given). The figure shows the very mechanism of the proof: each term overshoots the target and then falls back short of it, the odd sums go down, the even sums go up, and the limit is caught in the vise between the two: convergence by squeezing from both sides.

The error is bounded by the FIRST NEGLECTED TERM: $|S-S_{n}|\le\frac{1}{n+1}$. For $0.01$: $\frac{1}{n+1}\le 0.01\iff n\ge 99$: about a hundred terms. For $10^{-6}$: a million terms. The bound is simple and safe, but the convergence is SLOW: dividing the error by 100 costs 100 times more terms. That is the price of conditional convergence: the series $\sum\frac{(-1)^{n}}{n^{2}}$, by contrast, reaches $10^{-6}$ in about a thousand terms.

ABSOLUTE convergence: $\sum|a_{n}|$ converges (and it always implies the convergence of $\sum a_{n}$). CONDITIONAL: $\sum a_{n}$ converges but $\sum|a_{n}|$ diverges. Alternating harmonic series: converges (a), but its absolute values form the divergent harmonic series: CONDITIONAL. $\sum\frac{(-1)^{n}}{n^{2}}$: the absolute values form $\sum\frac{1}{n^{2}}$, a convergent p-series with $p=2$: ABSOLUTE. Hierarchy: absolute $\Rightarrow$ convergent, never the reverse.

Riemann's rearrangement theorem: by changing only the ORDER of the terms of a conditionally convergent series, you can make it converge to any real number, or diverge. The alternating harmonic series, reordered, can “equal” 5, or $-\pi$, or infinity: its sum of $\ln 2$ depends on the order of addition, which clashes with everything finite addition has taught us. Absolutely convergent series are immune: every permutation gives back the same sum. That is why absolute convergence is the safe notion: it alone allows you to handle an infinite sum like an ordinary one.

Geometric with ratio $r=-\frac{2}{3}$, $|r|=\frac{2}{3}<1$: ABSOLUTE convergence (the series of absolute values is the geometric series with ratio $\frac{2}{3}$). Sum: $\frac{1}{1-(-\frac{2}{3})}=\frac{1}{\frac{5}{3}}=\frac{3}{5}=0.6$. An alternating series can be handled with Leibniz, but when it is geometric, the exact sum comes for free: always recognize the geometric series first.

$(-1)^{n}\frac{n}{n+1}$: the general term does not tend to 0 (it oscillates between values close to $+1$ and $-1$): DIVERGENT by the divergence test: Leibniz does not apply ($b_{n}\to 1\ne 0$) and should not even be invoked. $\frac{n^{2}}{n^{4}+1}$: positive, compares with $\frac{n^{2}}{n^{4}}=\frac{1}{n^{2}}$: $\frac{n^{2}}{n^{4}+1}\le\frac{1}{n^{2}}$, p-series with $p=2$: CONVERGENT (absolutely, since it is positive).

$\cos(n\pi)=(-1)^{n}$: this is the alternating series $\sum\frac{(-1)^{n}}{\sqrt{n}}$ in disguise: first reflex, unmask the alternation. Absolute values: $\sum\frac{1}{\sqrt{n}}$, p-series with $p=\frac{1}{2}$: diverges: no absolute convergence. Leibniz: $\frac{1}{\sqrt{n}}$ decreases to 0: the series converges. Verdict: CONDITIONAL convergence: the subtlest case, which requires BOTH checks, and the one exams love.

$\frac{3^{n}}{n4^{n}}=\frac{1}{n}\left(\frac{3}{4}\right)^{n}$: ratio test: $\frac{n}{n+1}\cdot\frac{3}{4}\to\frac{3}{4}<1$: ABSOLUTE convergence (direct comparison with the geometric series $\left(\frac{3}{4}\right)^{n}$ works too: the $\frac{1}{n}$ only helps in the right direction). General strategy, in order: 1) does the term tend to 0? if not, divergence, done; 2) is the series geometric or telescoping? exact sum; 3) factorials or $c^{n}$? ratio test; 4) powers of $n$? p-series and comparison; 5) alternation? absolute convergence first, then Leibniz. Five questions, in that order: that is the algorithm that solves the exam sheet.

Root test (Cauchy): let $L=\lim_{n\to\infty}\sqrt[n]{\left|a_{n}\right|}$. If $L<1$ the series converges absolutely; if $L>1$ (or $L=\infty$) it diverges; if $L=1$ the test is inconclusive. Application: $\sqrt[n]{\left(\frac{n}{2n+1}\right)^{n}}=\frac{n}{2n+1}\to\frac{1}{2}<1$: the series CONVERGES absolutely. The root simply cancels the exponent $n$, which is exactly why it is the right tool here. The ratio test would be painful because you would have to form $\frac{(n+1)^{n+1}(2n+1)^{n}}{(2n+3)^{n+1}n^{n}}$ and untangle this quotient of $n$th powers with different exponents, which requires going through the logarithm and several intermediate limits. Rule of choice: as soon as the general term is an overall $n$th power, the root is immediate.

The term is $a_{n}=\frac{\left(1+\frac{1}{n}\right)^{n^{2}}}{2^{n}}$, with an exponent $n^{2}$ that rules out any direct treatment. $n$th root: $\sqrt[n]{a_{n}}=\frac{\left(1+\frac{1}{n}\right)^{n}}{2}$. Now $\left(1+\frac{1}{n}\right)^{n}\to e$, so $L=\frac{e}{2}\approx 1.359>1$: the series DIVERGES. Numerical check: for $n=50$, the root is already $1.346$, well above 1. The beauty of the question is that the answer hinges on comparing $e$ with 2: had the denominator been $3^{n}$, you would have obtained $\frac{e}{3}\approx 0.906<1$ and the series would have converged. The whole behaviour of the series is decided by two decimals of the number $e$, a reminder that the boundary $L=1$ really is a boundary and not a blurry zone.