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You're given this series: $$\sum_{n=1}^\infty e^{(1-\cos\left(\frac{1}n\right))} -1$$

We know that $e^{(1-\cos\left(\frac{1}n\right))} \ge 1$ since $(1-\cos\left(\frac{1}n\right))\ge 0$ then $\sum_{n=1}^\infty e^{(1-\cos\left(\frac{1}n\right))} -1\ge 0 $, so:

$$\sum_{n=1}^\infty e^{(1-\cos\left(\frac{1}n\right))} -1 \ge\sum_{n=1}^\infty -1$$

But $\sum_{n=1}^\infty -1$ diverges, so the original series diverge as well by the comparison test.
I tried this series in wolfram, and it said this series actually converge. What am I doing wrong?

Matam
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  • How does $i$ relate to $n$? – Barry Cipra Aug 01 '16 at 21:01
  • It's supposed to be n everywhete. My bad! – Matam Aug 01 '16 at 21:02
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    I've gone through and fixed some of the weird mathjax. You can check the general help guide to learn some more about mathjax; a lot of the things, like writing individual characters as superscripts, too many dollar signs, and parentheses vs. brackets should be fixed. –  Aug 01 '16 at 21:07
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    $\sum\ge-\infty$ doesn't allow you to conclude anything. –  Aug 01 '16 at 21:07
  • Also, you somehow went from $e^{(1-\cos(1/n))}\geq 1$ to $(e^{(1-\cos(1/n))}-1)\geq 0$ to $(e^{(1-\cos(1/n))}-1)\geq -1$ at some point - you inadvertently subtracted 1 from the right hand side twice. – Steven Stadnicki Aug 01 '16 at 21:44

4 Answers4

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You applied the comparison test incorrectly: There is usually some assumption like $b_n \ge a_n \ge 0$, and divergence of $\sum a_n$ leads to divergence of $\sum b_n$. After all, $0 \ge -1$, $\sum_{n = 1}^{\infty} -1$ diverges, and $\sum_{n = 1}^{\infty} 0$ clearly converges.

For an approach to the problem: Using Taylor series, you can verify that

$$e^{1 - \cos(1/n)} - 1 = e^{1/2n^2 + O(1/n^4)} - 1 = \frac 1 {2n^2} + O\left(\frac 1 {n^4}\right)$$ and use that $\sum_{n = 1}^{\infty} 1/n^p$ converges when $p > 1$.

  • So to use the comparison test as I did, I need both series, the one I check and the one I compare to, to be $\ge 0$, right? Thanks. – Matam Aug 01 '16 at 21:05
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    Yes, depending on exactly what form of the test you're using. –  Aug 01 '16 at 21:05
  • Why is this true: $$e^{1/2n^2 + O(1/n^4)} - 1 = \frac 1 {2n^2} + O\left(\frac 1 {n^4}\right)$$ – Matam Aug 01 '16 at 22:44
  • Take the Taylor series for $e^x = 1 + x + O(x^2)$. –  Aug 01 '16 at 23:03
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Let us take for granted that for $x<1$,

$$e^x\le\frac1{1-x}.$$

Then

$$\sum_{n=1}^\infty e^{x_n}-1\le\sum_{n=1}^\infty\frac {x_n}{1-x_n}$$

and with $x_n=1-\cos(1/n)=2\sin^2(1/2n)$, $$\sum_{n=1}^\infty e^{2\sin^2(\frac1{2n})}-1\le\sum_{n=1}^\infty\frac {2\sin^2(\frac1{2n})}{1-2\sin^2(\frac1{2n})}.$$

Then with $1/{4n}\le\sin1/{2n}\le1/{2n}$, $$\sum_{n=1}^\infty e^{2\sin^2(\frac1{2n})}-1\le\frac12\sum_{n=1}^\infty\frac {\frac2{4n^2}}{1-\frac2{16n^2}}\le\sum_{n=1}^\infty\frac1{4n^2}.$$

By convergence of the last series, the given series is bounded. As its terms are positive, it does converge.

  • @BarryCipra: typo fixed. –  Aug 01 '16 at 21:36
  • If you are saying $${2\sin^2({1\over2n})\over1-2\sin^2({1\over2n})}\le{1\over2}{{2\over4n^2}\over1-{2\over16n^2}}$$ I don't think that's right. That reduces to $${\sin^2x\over1-2\sin^2x}\le{x^2\over2-x^2}$$ with $x={1\over2n}$. – Barry Cipra Aug 01 '16 at 21:50
  • @BarryCipra: this is bounded above by $2x^2$ in a sufficient range. –  Aug 02 '16 at 06:25
  • I'm not questioning the final result, just the justification for a particular step. I don't see how you got from the sum with the sines in the penultimate inequality to the (middle) sum without them in the final displayed expression. It seems to me there should be a $2$ in front of that middle sum instead of a $1\over2$ (and then just a $1\over n^2$ in the final sum). – Barry Cipra Aug 02 '16 at 13:12
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I thought it would be instructive to present a way forward that relies on elementary inequalities only.

PRIMER $1$:

In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function satisfies the inequality

$$\bbox[5px,border:2px solid #C0A000]{e^x\le \frac{1}{1-x}} \tag 1$$

for $x<1$.

Using $(1)$, we can assert that for $n>1$

$$0\le e^{1-\cos(1/n)}-1\le \frac{2\sin^2\left(\frac1{2n}\right)}{\cos(1/n)} \tag 2$$

PRIMER $2$:

Now, it is easy to show from elementary geometry that $x\cos (x)\le \sin(x)\le x$ for $0\le x \le \pi/2$. And from this set of inequalities we have that

$$\begin{align}\sin(x)&\le x \tag 3\\\\ \cos(x)&\ge \sqrt{1-x^2} \tag 4 \end{align}$$

for $0\le x \le \pi/2$.

Using $(3)$ and $(4)$ in $(2)$ reveals for $n>1$

$$0 \le e^{1-\cos(1/n)}-1\le \frac{1}{2n^2\sqrt{1-(1/n)^2}}\le \frac{1}{n^2}$$

Since the series $\sum_{n=1}^{\infty}\frac{1}{n^2}=\pi^2/6$ converges, the series of interest does likewise. And we are done.

Community
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Mark Viola
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You could have used the ratio test to $$u_n=e^{1-\cos\left(\frac{1}n\right)} -1$$ $$\frac{u_{n+1}}{u_n}=\frac{e^{1-\cos \left(\frac{1}{n+1}\right)}-1}{e^{1-\cos \left(\frac{1}{n}\right)}-1}$$ and using Taylor series for large values of $n$ (just as T. Bongers answered) to get $$\frac{u_{n+1}}{u_n}=1-\frac{2}{n}+O\left(\frac{1}{n^2}\right)$$ Similarly, using Raabe's test $$n\left( \frac{u_n}{u_{n+1}}-1\right)=n \left(\frac{e^{1-\cos \left(\frac{1}{n}\right)}-1}{e^{1-\cos \left(\frac{1}{n+1}\right)}-1}-1\right)=2+\frac{1}{n}+O\left(\frac{1}{n^2}\right)$$

Claude Leibovici
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