# Can't tell whether $\sum_{n=1}^\infty e^{1-\cos(\frac{1}n)} -1$ converges or not

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?

• How does $i$ relate to $n$? – Barry Cipra Aug 1 '16 at 21:01
• It's supposed to be n everywhete. My bad! – Matam Aug 1 '16 at 21:02
• 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. – user296602 Aug 1 '16 at 21:07
• $\sum\ge-\infty$ doesn't allow you to conclude anything. – Yves Daoust Aug 1 '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 1 '16 at 21:44

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 1 '16 at 21:05
• Yes, depending on exactly what form of the test you're using. – user296602 Aug 1 '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 1 '16 at 22:44
• Take the Taylor series for $e^x = 1 + x + O(x^2)$. – user296602 Aug 1 '16 at 23:03

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. – Yves Daoust Aug 1 '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 1 '16 at 21:50
• @BarryCipra: this is bounded above by $2x^2$ in a sufficient range. – Yves Daoust Aug 2 '16 at 6:25
• 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 2 '16 at 13:12

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.

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 $$n\left( \frac{u_n}{u_{n+1}}-1\right)=n \left(\frac{e^{1-\cos \left(\frac{1}{n}\right)}-1}{e^{1-\cos \left(\frac{1}{n+1}\right)}-1}-1\right)=2+\frac{1}{n}+O\left(\frac{1}{n^2}\right)$$