Check whether the series is convergent?
$\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}\tan(\frac{1}{n})$
$\endgroup$ 12 Answers
$\begingroup$HINT: Use the comparison test.
$\endgroup$ $\begingroup$$$\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}\tan(\frac{1}{n}) = \sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}\frac{\sin(\frac{1}{n})}{\cos(\frac{1}{n})}$$ note.. $$\lim_{x\to0}\frac{\sin x}{x}=1 \implies \sin(x)\approx x$$ for small $x$ values.. thus.. for large $n$'s $$\sin\left(\frac1n\right)\approx \frac1n$$ thus... $$\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}\frac{\sin(\frac{1}{n})}{\cos(\frac{1}{n})}\approx \sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}\frac{ \frac{1}{n}}{\cos(\frac{1}{n})}$$ for large $n$'s $$\cos\left(\frac1n\right)\approx 1$$ thus... $$\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}\frac{\sin(\frac{1}{n})}{\cos(\frac{1}{n})}\approx \sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}\frac{ \frac{1}{n}}{ 1} = \sum \frac{1}{n^{3/2}}$$ which clearly converges by recognizing it as a p-series, or integral test.. etc....
$\endgroup$
