I am trying to determine if $ \sum_{n = 1}^{\infty} (-1)^{n + 1} \left( 1 - n \log \left( \frac{n + 1}{n} \right) \right) $ converges using an alternating series test. The test in question requires me to prove $ 1 - n \log \left( \frac{n + 1}{n} \right) $ is decreasing and that $ \lim_{n \to \infty} \left( 1 - n \log \left( \frac{n + 1}{n} \right) \right) = 0 $ to prove this series is convergent.

My intutition says that this series is convergent because $ n \log \left( \frac{n + 1}{n} \right) $ will tend towards 1 as n goes to infinity (due to the definition of e). I am having trouble proving the sequence $ 1 - n \log \left( \frac{n + 1}{n} \right) $ is decreasing.

I set up the inequality $ 1 - (n + 1) \log \left( \frac{n + 2}{n + 1} \right) \leq 1 - n \log \left( \frac{n + 1}{n} \right) $ but I feel like I am stuck on some simple algebra. Any hints for proving the sequence is nonincreasing? Or am I just wrong?

EDIT: Could this be done using absolute convergence? EDIT2: I am seeing some really great answers, but I am trying to prove this without calculus. (No derivatives or Taylor series expansions.)

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    $\begingroup$ Did you notice that $\lim_{n\to\infty} n \ln \left( \frac{n + 1}{n} \right)=\lim_{n\to\infty} \ln \left( 1+ \frac{1}{n} \right)^n=\ln e=1$? $\endgroup$ – draks ... Jul 25 '12 at 8:19
  • $\begingroup$ hint: expand for example in Taylor series in $\frac 1n$. $\endgroup$ – Raymond Manzoni Jul 25 '12 at 8:21
  • $\begingroup$ Have you tried to consider the function $f \colon x \mapsto 1-x\log{\frac{x+1}{x}}$? It is differentiable, so you can study the sign of $f'$ in order to get information on the monotonicity... $\endgroup$ – Romeo Jul 25 '12 at 8:22
  • $\begingroup$ @draks - You should post that as an answer. In combination with the second sentence of the body of the question, this is a complete solution. $\endgroup$ – user22805 Jul 25 '12 at 8:25
  • $\begingroup$ @DavidWallace Huh? This leaves out the monotonicity. $\endgroup$ – Did Jul 25 '12 at 8:41

Using the Taylor expansion of $\log(1+x)$ we see that $$ 1-n\log\Bigl(\frac{n+1}{n}\Bigr)=1-n\Bigl(\frac{1}{n}-\frac{1}{2\,n^2}+O(n^{-3})\Bigr)=\frac{1}{2\,n}+a_n,\quad a_n=O(n^{-2}). $$ Then $$ \sum_{n=1}^\infty(-1)^{n+1}\Bigl(1-n\log\Bigl(\frac{n+1}{n}\Bigr)\Bigr)=\sum_{n=1}^\infty(-1)^{n+1}\frac{1}{2\,n}+\sum_{n=1}^\infty(-1)^{n+1}a_n. $$ The first series on the right hand side converges by Leibniz's test, and the second is absolutely convergent since $|(-1)^{n+1}a_n|\le C\,n^{-2}$ for some constant $C>0$.


Look at $x \mapsto 1-x\log\left(\frac{x+1}{x}\right)$. Its second derivative is positive so its first derivative is increasing. It first derivative tends to $-\infty$ in when $x$ goes to zero and to $0$ when $x$ goes to $\infty$, thus it is negative, and the initial function is decreasing. Consider its restriction to $\mathbb{N}$ then.


Notice that $$\lim_{n\to\infty} n \ln\left( \frac{n + 1}{n} \right)=\lim_{n\to\infty} \ln\left( 1+ \frac{1}{n} \right)^n=\ln e=1.$$

So $\lim_{n \to \infty} \left( 1 - n \log \left( \frac{n + 1}{n} \right) \right) = 0$ as required and we are left with monotonicity:

$$ 1- (n + 1) \log \left( \frac{n + 2}{n + 1} \right) \leq 1- n \log \left( \frac{n + 1}{n} \right)\\ \log \left( \frac{n + 2}{n + 1} \right)^{n+1} \geq \log \left( \frac{n + 1}{n} \right)^n\\ \left( \frac{n + 2}{n + 1} \right)^{n+1} \geq \left( \frac{n + 1}{n} \right)^n\\ \left( 1+ \frac1{n+1} \right)^{n+1} \Biggr/ \left(1+ \frac1n \right)^n \geq 1\\ \left(\left( 1+ \frac1{n+1} \right) \Biggr/ \left(1+ \frac1n \right) \right)^n \geq \frac{n+1}{n+2}\\ $$ Now use $ \left(\left( 1+ \frac1{n+1} \right) \Biggr/ \left(1+ \frac1n \right) \right)^n \geq \left(\left( 1+ \frac1{n+1} \right) \Biggr/ \left(1+ \frac1n \right) \right)=\frac{n(n+2)}{(n+1)^2}$ to conclude: $$ \frac{n(n+2)}{(n+1)^2} \geq \frac{n+1}{n+2}\\ \frac{n(n+2)^2}{(n+1)^3}=\frac{n^3+4n^2+4n}{n^3+3n^2+3n+1} \geq 1\; ,\\ $$ for $n\geq 1$, which is not elegant, I know.

  • $\begingroup$ Note that that's not enough for the alternating series to converge. The function needs to go to zero monotonically. It's easy to come up with an example of a function going to zero not monotonically that makes the alternating series divergent. $\endgroup$ – bartgol Jul 25 '12 at 8:46
  • $\begingroup$ But $(1 + \frac{1}{n})^{n}$ increases monotonically to $e,$ though that takes a little work and thought. So (the absolute values of) the terms in the quantity being summed do decrease monotonically, and it is OK in this case. $\endgroup$ – Geoff Robinson Jul 25 '12 at 9:03
  • $\begingroup$ @GeoffRobinson Would you mind having a look at my (not very elegant) proof for the monotonicity? $\endgroup$ – draks ... Jul 25 '12 at 14:17
  • $\begingroup$ Replace $+2n$ by $+4n$ at the very end. $\endgroup$ – Did Jul 25 '12 at 14:55
  • $\begingroup$ It looks correct, apart from the 2n-4n typo which did points out. A few more words might help though :) $\endgroup$ – Geoff Robinson Jul 25 '12 at 15:22

Recall that for all $x>0$ $$ 1 - \frac1x \leq \log x \leq x-1 \tag{1} $$

If $x=\frac{n+1}{n}$, then $x>1$ and $n=\frac{1}{x-1}$. Multiplying $({1})$ by $-n=\frac{1}{1-x}<0$ and then adding $1$ we have $$ 1-n+\frac{n}{x}\geq1-n\log x \geq 1-nx+n $$ or $$ 1-n+\frac{n^2}{n+1}\geq1-n\log \frac{n+1}{n} \geq 1-n\frac{n+1}{n}+n=0 $$ A more explicit estimate would be $$ 1-\frac{1}{n+1}\geq1-n\log \frac{n+1}{n} \geq 0 \tag{2} $$

By the criterion of convergence of alternating series we have that the series $ \sum_{n = 1}^{\infty} (-1)^{n + 1} \left( 1 - \frac{1}{n+1}\right) $ converges. By the criterion of comparison for alternating series, it follows from $(2)$ that the series $$ \sum_{n = 1}^{\infty} (-1)^{n+1} \left( 1 - n \log \left( \frac{n + 1}{n} \right) \right) $$ converges.


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