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?
 A: 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 $p > 1$.
A: 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.
A: 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.
A: 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 $$n\left( \frac{u_n}{u_{n+1}}-1\right)=n \left(\frac{e^{1-\cos \left(\frac{1}{n}\right)}-1}{e^{1-\cos
   \left(\frac{1}{n+1}\right)}-1}-1\right)=2+\frac{1}{n}+O\left(\frac{1}{n^2}\right)$$
