Corrected exercises: advanced convergence tests and classic pitfalls, 201-SN3-RE (201-NYB)

The series $\sum a_{n}$ converges if and only if: for every $\varepsilon>0$, there is an index $N$ such that for all $q>p\geq N$, $\left|\sum_{n=p+1}^{q}a_{n}\right|<\varepsilon$. In other words, blocks of consecutive terms become arbitrarily small, however long they are.

Taking $q=p+1$, the sum reduces to the single term $a_{p+1}$, and the condition becomes $\left|a_{p+1}\right|<\varepsilon$ as soon as $p\geq N$: this is exactly the definition of $a_{n}\to 0$. The divergence test is therefore a simple corollary of the Cauchy criterion, which explains why it is necessary but not sufficient: it only uses blocks of length 1.

Take $q=2p$. Each of the $p$ terms of $\sum_{n=p+1}^{2p}\frac{1}{n}$ is greater than or equal to $\frac{1}{2p}$, so $\sum_{n=p+1}^{2p}\frac{1}{n}\geq p\cdot\frac{1}{2p}=\frac{1}{2}$. These blocks never become small, whatever $p$ is: the Cauchy criterion fails with $\varepsilon=\frac{1}{2}$, so the harmonic series DIVERGES. (Numerically, these blocks actually tend to $\ln 2\approx 0.693$, well above $\frac{1}{2}$.)

For $n\geq 2$, $\frac{1}{n^{2}}\leq\frac{1}{n(n-1)}=\frac{1}{n-1}-\frac{1}{n}$. Summing from $p+1$ to $q$, the sum telescopes: $\sum_{n=p+1}^{q}\frac{1}{n^{2}}\leq\frac{1}{p}-\frac{1}{q}<\frac{1}{p}$. Given $\varepsilon>0$, it is enough to take $N>\frac{1}{\varepsilon}$ for every block to be smaller than $\varepsilon$. The series CONVERGES.

It is valuable because it characterizes convergence without ever mentioning the value of the sum: it is the tool used to prove the other tests, and it is what ties convergence to the completeness of $\mathbb{R}$. In practice it is not used, because ALL the blocks must be controlled at once, which is much harder than applying a comparison. It therefore serves mainly as a proof tool, and to establish divergence as in c.

Compute $2^{n}a_{2^{n}}=2^{n}\cdot\dfrac{1}{\left(2^{n}\right)^{p}}=2^{n(1-p)}=\left(2^{1-p}\right)^{n}$. It is a geometric series with common ratio $r=2^{1-p}$, which converges if and only if $r<1$.

$2^{1-p}<1$ is equivalent to $1-p<0$, that is, $p>1$. The p-series converges if and only if $p>1$. Condensation therefore proves the p-series test in three lines, without any integral: it reduces everything to a geometric series, whose behaviour has been known since the start of the chapter.

$2^{n}a_{2^{n}}=\dfrac{2^{n}}{2^{n}\ln\left(2^{n}\right)}=\dfrac{1}{n\ln 2}$. This is the harmonic series up to a constant factor, which DIVERGES. So $\sum\frac{1}{n\ln n}$ diverges, even though its general term is much smaller than $\frac{1}{n}$.

$2^{n}a_{2^{n}}=\dfrac{2^{n}}{2^{n}\left(n\ln 2\right)^{\beta}}=\dfrac{1}{(\ln 2)^{\beta}}\cdot\dfrac{1}{n^{\beta}}$. Up to a constant, this is a p-series with exponent $\beta$: it converges if and only if $\beta>1$. This is exactly Bertrand's test, and the proof also explains WHY the logarithm plays the role of the variable again: condensation turns $\ln 2^{n}$ into $n\ln 2$.

The decreasing hypothesis is essential to bound each block by its first and last terms. Idea: group the terms into blocks $\left[2^{k},2^{k+1}\right)$, which contain $2^{k}$ terms. Since the sequence decreases, each term lies between $a_{2^{k+1}}$ and $a_{2^{k}}$, so the block lies between $2^{k}a_{2^{k+1}}$ and $2^{k}a_{2^{k}}$. Summing over $k$, the whole series lies between $\frac{1}{2}\sum 2^{k+1}a_{2^{k+1}}$ and $\sum 2^{k}a_{2^{k}}$: the two series therefore behave the same way. Without the decreasing hypothesis the bounds collapse, and the theorem is false.

$a_{2}=2^{-3}=\frac{1}{8}=0.125$, $a_{3}=2^{-2}=\frac{1}{4}=0.25$, $a_{4}=2^{-5}=\frac{1}{32}$, $a_{5}=2^{-4}=\frac{1}{16}$. The ratios: $\frac{a_{3}}{a_{2}}=2$ and $\frac{a_{4}}{a_{3}}=\frac{1}{8}$. The sequence is not even decreasing: it zigzags.

$\dfrac{a_{n+1}}{a_{n}}=2^{-(n+1)-(-1)^{n+1}+n+(-1)^{n}}=2^{-1+2(-1)^{n}}$. For even $n$ this equals $2^{1}=2$, for odd $n$ it equals $2^{-3}=\frac{1}{8}$. The ratio oscillates forever between 2 and $\frac{1}{8}$: it has no limit, and the ratio test is INCONCLUSIVE. It does not even conclude divergence, although the ratio exceeds 1 every other time.

$\sqrt[n]{a_{n}}=2^{\frac{-n-(-1)^{n}}{n}}=2^{-1-\frac{(-1)^{n}}{n}}$. Since $\frac{(-1)^{n}}{n}\to 0$, the limit is $2^{-1}=\frac{1}{2}<1$: the root test concludes CONVERGENCE. (The sum is in fact exactly $\frac{3}{2}$, splitting the even and odd indices into two geometric series.)

Theorem: if $\left|\frac{a_{n+1}}{a_{n}}\right|$ has a limit $L$, then $\sqrt[n]{|a_{n}|}$ has the same limit $L$. The converse is false, as the example above shows. The root test is therefore STRICTLY STRONGER: whenever the ratio test concludes, the root test reaches the same conclusion, and it sometimes concludes when the ratio test is silent. In practice the ratio test still remains the first reflex, because ratios of factorials simplify far more easily than their $n$th roots.

The harmonic series $\sum\frac{1}{n}$: the ratio tends to 1 and so does the $n$th root, since $\sqrt[n]{\frac{1}{n}}=e^{-\frac{\ln n}{n}}\to 1$. Both tests fail together, and that is no accident: both implicitly compare the series with a geometric series, and no geometric series can mimic a polynomial decrease. You then conclude with the p-series test, by comparison or with the integral test.

$\dfrac{a_{n+1}}{a_{n}}=\left(\dfrac{n}{n+1}\right)^{p}=\left(1+\dfrac{1}{n}\right)^{-p}=1-\dfrac{p}{n}+o\!\left(\dfrac{1}{n}\right)$ by the Taylor expansion. So $\alpha=p$, and the test gives convergence for $p>1$ and divergence for $p<1$: exactly the p-series test. (The case $p=1$ escapes the test, as always on the boundary.)

$a_{1}=\frac{1}{2}$, $a_{2}=\frac{1\cdot 3}{2\cdot 4}=\frac{3}{8}=0.375$, $a_{3}=\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}=\frac{15}{48}=\frac{5}{16}=0.3125$. The ratio simplifies drastically, since all the common factors cancel: $\frac{a_{n+1}}{a_{n}}=\frac{2n+1}{2n+2}$.

$\dfrac{2n+1}{2n+2}\to 1$: the ratio test is inconclusive. Expand: $\frac{2n+1}{2n+2}=1-\frac{1}{2n+2}=1-\frac{1}{2n}+o\!\left(\frac{1}{n}\right)$, so $\alpha=\frac{1}{2}$. Since $\alpha<1$, the series DIVERGES. Numerical check: $n\left(1-\frac{a_{n+1}}{a_{n}}\right)$ equals 0.490 for $n=50$ and 0.499 for $n=500$, which confirms $\alpha=\frac{1}{2}$.

If $a_{n}\sim\frac{1}{\sqrt{\pi n}}=\frac{1}{\sqrt{\pi}}n^{-1/2}$, then $\sum a_{n}$ behaves like a p-series with exponent $\frac{1}{2}\leq 1$: it diverges, which confirms c. Next, $a_{n}^{2}\sim\frac{1}{\pi n}$, exponent 1: $\sum a_{n}^{2}$ DIVERGES too. Finally, $a_{n}^{3}\sim\frac{1}{(\pi n)^{3/2}}$, exponent $\frac{3}{2}>1$: $\sum a_{n}^{3}$ CONVERGES.

Because the ratio test only keeps the LIMIT of the ratio and throws away all the information on how that limit is reached. Yet between $1-\frac{1}{2n}$ and $1-\frac{2}{n}$, the limit is the same but the decrease of $a_{n}$ is radically different, like $n^{-1/2}$ in one case and like $n^{-2}$ in the other. Raabe's test reads precisely the $\frac{1}{n}$ term of the expansion, that is, what the ratio test had neglected.

$\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$. The partial sum telescopes: $S_{N}=1-\frac{1}{N+1}\to 1$. The sum is exactly 1.

$\frac{1}{n(n+2)}=\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right)$. Since the shift is 2, the terms cancel two indices apart, and TWO terms REMAIN at the start: $S_{N}=\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{N+1}-\frac{1}{N+2}\right)\longrightarrow\frac{1}{2}\cdot\frac{3}{2}=\frac{3}{4}$. The general rule: a shift of $k$ leaves $k$ terms on each side.

$\frac{1}{n(n+1)(n+2)}=\frac{1}{2}\left(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right)$. This is again a telescoping sum, on the sequence $u_{n}=\frac{1}{n(n+1)}$: the sum equals $\frac{1}{2}u_{1}=\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$.

Use a common denominator: $\frac{1}{n^{2}}-\frac{1}{(n+1)^{2}}=\frac{(n+1)^{2}-n^{2}}{n^{2}(n+1)^{2}}=\frac{2n+1}{n^{2}(n+1)^{2}}$, which is indeed the stated identity. The sum therefore telescopes to $\frac{1}{1^{2}}=1$.

$1-\frac{1}{n^{2}}=\frac{n^{2}-1}{n^{2}}=\frac{(n-1)(n+1)}{n^{2}}$, so $\ln\left(1-\frac{1}{n^{2}}\right)=\ln(n-1)+\ln(n+1)-2\ln n=\left[\ln(n+1)-\ln n\right]-\left[\ln n-\ln(n-1)\right]$. This is the difference of two consecutive terms of the sequence $v_{n}=\ln(n+1)-\ln n$, so the sum from $2$ to $N$ equals $v_{N}-v_{1}=\ln\frac{N+1}{N}-\ln 2\to-\ln 2$. The sum is $-\ln 2\approx-0.693$. Note in passing that all the terms are negative, which is consistent.

$\dfrac{b_{n}}{a_{n}}=1+\dfrac{1/n}{(-1)^{n}/\sqrt{n}}=1+\dfrac{(-1)^{n}}{\sqrt{n}}\longrightarrow 1$. The two sequences are therefore indeed equivalent in the usual sense.

$\sum a_{n}$ is an alternating series whose absolute value $\frac{1}{\sqrt{n}}$ decreases to 0: it CONVERGES (conditionally). On the other hand, $\sum b_{n}=\sum a_{n}+\sum\frac{1}{n}$, the sum of a convergent series and the divergent harmonic series: it DIVERGES. Two series with equivalent terms thus behave in OPPOSITE ways here, which is impossible for positive terms.

The limit comparison test requires the terms to have a constant sign, at least from some index on. Here $a_{n}$ changes sign at every index. The equivalence $b_{n}\sim a_{n}$ controls the ratio, but not the DIFFERENCE: $b_{n}-a_{n}=\frac{1}{n}$ is negligible compared with $a_{n}\sim n^{-1/2}$ (in absolute value), and yet its series diverges. With positive terms such an accident cannot happen, because convergence is then equivalent to the partial sums simply being bounded above.

No. The Taylor expansion gives $\ln\left(1+\frac{(-1)^{n}}{\sqrt{n}}\right)=\frac{(-1)^{n}}{\sqrt{n}}-\frac{1}{2n}+O\!\left(\frac{1}{n^{3/2}}\right)$. The first term gives a convergent series and so does the third (a p-series with exponent $\frac{3}{2}$), but the middle term, $-\frac{1}{2n}$, produces a divergent series. The series therefore DIVERGES, and its partial sums behave like $-\frac{1}{2}\ln N$. This is the typical counterexample: the equivalent was correct, the conclusion drawn from it was wrong.

On a series whose terms change sign, NEVER reason with equivalents. Proceed in this order: first test for absolute convergence, which settles the matter if $\sum|a_{n}|$ converges; otherwise, expand the general term to an order high enough to separate an alternating part, handled by the alternating series test (Leibniz), from a remainder of constant sign, handled by the p-series test; conclude on the SUM of these pieces. This is exactly the method of question d.

To first order, $\sin\frac{1}{n}\sim\frac{1}{n}$, so the difference is equivalent to $0$: the information is lost, and a zero equivalent says nothing. Push the expansion further: $\sin u=u-\frac{u^{3}}{6}+O(u^{5})$ with $u=\frac{1}{n}$, hence $\sin\frac{1}{n}-\frac{1}{n}=-\frac{1}{6n^{3}}+O\!\left(\frac{1}{n^{5}}\right)\sim-\frac{1}{6n^{3}}$. The terms have a constant sign (negative) and are comparable to $n^{-3}$: the series CONVERGES absolutely. Check at $n=100$: the difference equals $-1.6667\times 10^{-7}$, and $-\frac{1}{6\cdot 10^{6}}$ equals $-1.6667\times 10^{-7}$.

$\ln(1+u)=u-\frac{u^{2}}{2}+O(u^{3})$, so $\frac{1}{n}-\ln\left(1+\frac{1}{n}\right)\sim\frac{1}{2n^{2}}$. Positive terms, exponent $2>1$: the series CONVERGES. At $n=100$ the difference equals $4.967\times 10^{-5}$, close to $\frac{1}{2\cdot 10^{4}}=5\times 10^{-5}$.

$e^{u}=1+u+\frac{u^{2}}{2}+O(u^{3})$, so $e^{1/n}-1-\frac{1}{n}\sim\frac{1}{2n^{2}}$: the series CONVERGES, for the same reason as in b. At $n=100$: $5.017\times 10^{-5}$.

No, it is not enough: a term that tends to 0 is necessary but never sufficient. Expand: $\left(1+\frac{1}{n}\right)^{n}=\exp\left(n\ln\left(1+\frac{1}{n}\right)\right)=\exp\left(1-\frac{1}{2n}+O\!\left(\frac{1}{n^{2}}\right)\right)=e\left(1-\frac{1}{2n}+O\!\left(\frac{1}{n^{2}}\right)\right)$, so the term equals $-\frac{e}{2n}+O\!\left(\frac{1}{n^{2}}\right)\sim-\frac{e}{2n}$. Terms of constant sign, equivalent to a multiple of $\frac{1}{n}$: the series DIVERGES. At $n=1000$ the term equals $-1.3579\times 10^{-3}$, against $-\frac{e}{2000}=-1.3591\times 10^{-3}$.

You must expand as far as the FIRST NONZERO TERM. That term provides the equivalent you are looking for, and it alone decides the behaviour; everything else is negligible compared with it. On a series whose terms change sign, you must go one step further still, until every term whose series diverges has been isolated, as in the previous exercise.

Abel's test: if $(a_{n})$ is a positive DECREASING sequence that tends to 0, and if the partial sums $B_{n}=\sum_{k=1}^{n}b_{k}$ are BOUNDED, then $\sum a_{n}b_{n}$ converges. Special case: with $b_{n}=(-1)^{n}$, the partial sums are alternately $-1$ and $0$, so they are bounded by 1; this is exactly the alternating series test. Leibniz's test is therefore nothing but Abel's test with the simplest possible oscillating part.

The geometric sum with ratio $e^{i}\neq 1$ equals $\sum_{k=1}^{n}e^{ik}=e^{i}\,\frac{e^{in}-1}{e^{i}-1}$, whose modulus is bounded by $\dfrac{|e^{in}-1|}{|e^{i}-1|}\leq\dfrac{2}{|e^{i}-1|}=\dfrac{2}{2\left|\sin\frac{1}{2}\right|}=\dfrac{1}{\left|\sin\frac{1}{2}\right|}\approx 2.086$. Since $\sum_{k=1}^{n}\sin k$ is the imaginary part of this sum, it is bounded by the same constant. (Numerically, the largest value observed up to $n=3000$ is about 1.958: the bound is correct and fairly tight.)

Apply Abel's test with $a_{n}=\frac{1}{n}$, positive, decreasing and with limit zero, and $b_{n}=\sin n$, whose partial sums are bounded by b. The test gives the CONVERGENCE of $\sum\frac{\sin n}{n}$. Its sum is in fact $\frac{\pi-1}{2}\approx 1.0708$.

No. We have $|\sin n|\geq\sin^{2}n=\frac{1-\cos 2n}{2}$, so $\frac{|\sin n|}{n}\geq\frac{1}{2}\cdot\frac{1}{n}-\frac{1}{2}\cdot\frac{\cos 2n}{n}$. The series $\sum\frac{\cos 2n}{n}$ converges by the same Abel argument, while $\sum\frac{1}{n}$ diverges. The lower bound is therefore the difference of a divergent series and a convergent series: it diverges, and by comparison $\sum\frac{|\sin n|}{n}$ DIVERGES. The series is CONDITIONALLY CONVERGENT.

For $\sum\frac{\sin n}{\sqrt{n}}$: yes, $\frac{1}{\sqrt{n}}$ still decreases to 0 and the oscillating part is unchanged, so the series converges (still conditionally). For $\sum\sin n$: no, because there is no decreasing factor any more; $a_{n}=1$ does not tend to 0. And indeed the series diverges, since its general term $\sin n$ does not tend to 0.

$c_{0}=a_{0}b_{0}$; $c_{1}=a_{0}b_{1}+a_{1}b_{0}$; $c_{2}=a_{0}b_{2}+a_{1}b_{1}+a_{2}b_{0}$. This is the rule for multiplying polynomials, where the coefficient of $x^{n}$ gathers all the products of indices adding up to $n$.

If $\sum a_{n}$ and $\sum b_{n}$ converge ABSOLUTELY, then $\sum c_{n}$ converges absolutely and its sum is the product of the two sums. (Mertens' theorem weakens the hypothesis: it is enough for one of the two series to converge absolutely and for the other one simply to converge.)

Both exponential series converge absolutely for all $x$ and $y$, so the hypothesis is satisfied. Compute $c_{n}=\sum_{k=0}^{n}\frac{x^{k}}{k!}\cdot\frac{y^{n-k}}{(n-k)!}=\frac{1}{n!}\sum_{k=0}^{n}\binom{n}{k}x^{k}y^{n-k}=\frac{(x+y)^{n}}{n!}$ by the binomial theorem. Summing: $e^{x}e^{y}=\sum\frac{(x+y)^{n}}{n!}=e^{x+y}$. Numerical check with $x=0.7$ and $y=0.4$: both sides equal 3.0041660239.

These series converge (they are alternating) but not absolutely, since $\sum\frac{1}{\sqrt{n+1}}$ is a p-series with exponent $\frac{1}{2}$. All the products $a_{k}a_{n-k}$ have the same sign $(-1)^{n}$, so no cancellation occurs and $\left|c_{n}\right|=\sum_{k=0}^{n}\frac{1}{\sqrt{(k+1)(n-k+1)}}$. Now $(k+1)(n-k+1)\leq\left(\frac{n}{2}+1\right)^{2}$ by the AM-GM inequality, so each term is at least $\frac{1}{\frac{n}{2}+1}$ and $\left|c_{n}\right|\geq\frac{n+1}{\frac{n}{2}+1}\to 2$. The general term does not tend to 0: the Cauchy product DIVERGES. Numerically, $\left|c_{200}\right|\approx 2.94$ and $\left|c_{800}\right|\approx 3.04$.