# Convergence/Divergence of the series $\sum\limits_{n=1}^{\infty}\tan(1/n)$

Trying to see if $$\sum\limits_{n=1}^{\infty}\tan(1/n)$$ converges or diverges. As $n \to \infty$, $\tan(1/n) \to 0$, so inconclusive. Ratio test was inconclusive, root test doesn't look good for this one, and it's clearly not an alternating series, so that leads to perhaps some comparison arguments. Not sure what to use for comparison, other than maybe $\sin(1/n)$, but I know nothing about that series.

Note that $$\lim_{n\to\infty}\frac{\tan(1/n)}{1/n} = \lim_{t\to0}\frac{\tan(t)}{t} \stackrel{L}{=} \lim_{t\to0}\frac{\sec^2(t)}{1} = 1$$ The limit comparision test then implies that either both $\sum1/n$ and $\sum\tan(1/n)$ converge or both diverge. But $\sum 1/n$ diverges so $\sum \tan(1/n)$ diverges.

• (+1) nice to know the importance of $\stackrel{L}{=}$ notation. – kaka Oct 21 '14 at 6:01

Hint: $\tan x\geq x$ for small $x\geq 0$.

• "for small nonnegative $x$" – Erick Wong Oct 21 '14 at 5:44
• Sure, modified for clarity (clearly here we deal with nonnegative quantities). – Milly Oct 21 '14 at 5:46

Hint: $$n \to + \infty ,\tan \left( {{1 \over n}} \right)\sim {1 \over n} \to divergent$$

Hint: The ratio $\frac{\tan(1/n)}{1/n}$ has limit $1$. Then use the Limit Comparison Test.

For small values of x, $tan(x)$ is approximately equal to x.

Try comparing this series to $\frac{1}{n}$. The test should show that your series diverges, since $\frac{1}{n}$ diverges.