# Convergence of $\sum_ {n=1}^\infty (1-n\sin\frac1n)^\alpha$ and $\sum_ {n=1}^\infty 2^n (\tan x)^{n^2}$

I was trying to solve a question of an entrance exam. I am having trouble in a particular type of problems. Please help me to solve.

(Actually my last 2 questions are also from these exam papers. I am sorry for this type of repetitions.I am only seeking help from MSE.)

The question basically asks to find the values of some parameter or variable for which the given infinite series converges. Examples of two such questions are given below.

1) For exactly which real values of $\alpha$ is the series $$\displaystyle \sum_ {n=1}^\infty (1-n\sin\frac1n)^\alpha$$ converges?

2) For which real numbers $x$ is the series $$\displaystyle \sum_ {n=1}^\infty 2^n (\tan x)^{n^2}$$ converges?

Now I can not understand how should I proceed for these kind of problems. What should be my strategy to handle these problems? Should I use test for convergence and how should I choose that?

Sorry for so many questions but I am really stuck. I cannot proceed with these problems.I apologise.

• Entrance exam at which place ? – Gabriel Romon Apr 29 '15 at 17:10
• @LeGrandDODOM PhD entrance test for Indian Institute of Technology Bombay – usermath Apr 29 '15 at 17:12
• Is it JEE, GATE, CEED or JAM ? – Gabriel Romon Apr 29 '15 at 17:18
• No, its their own qualifying exam to shortlist candidates (for interview) who have qualified GATE (or equivalent nation wide test) with a prescribed cutoff. – usermath Apr 29 '15 at 17:20
• I see, thank you. – Gabriel Romon Apr 29 '15 at 17:20

$1)$ $\sin(\frac{1}{n}) \sim \frac{1}{n} - \frac{1}{6n^3}$
$2)$ Use root test.
• First of all thnx for your help. For (2) $n$ th root of the term is $2(\tan x)^n$ and its limit if less than 1 then the series convergence. But how can I get $x$ from it? Can you please explain a little bit more. For (1) I am unable to understand your hint. It will be helful for me if you kindly give some more details. – usermath Apr 29 '15 at 17:11
• On the principal branch, $|\tan x| <1$ when $|x|< \pi/4$. Can you find the other intervals for which $\tan x <1|$? – Mark Viola Apr 29 '15 at 17:22
• Sure. For $2)$ the $n$th root is $2 |\tan x|^n$. So it depends on whether $|\tan x|$ is $< 1$, $>1$ or $=1$. In case it is $=1$, the limit is $2 >1$. In case $<1$, the limit is $0<1$. In case $>1$, that's $\infty > 1$. So you just have to study the function $\tan$ for these cases. E.g. $\tan x = 1 \iff x = \frac{\pi}{4} + k\pi$ or something (I don't remember the details). For $1)$, that is saying that $sin(\frac{1}{n})$ behaves just like that sequence for large enough $n$. So we can replace it by that expression to study our series. In which case it becomes a Riemann series. – user230734 Apr 29 '15 at 17:24
• @BolzWeir Thank you very much. I understand (2) now. I will find the intervals there $|\tan x|<1$. But I am still confused about (1). Can you please explain what Riemann series is and how should I proceed. – usermath Apr 29 '15 at 17:34
• You're welcome. Regarding $1)$, my method of working won't be quite rigorous. If you want rigour, you should probably wait for another user to come and fix things. But note that the result is true. Replacing $\sin(\frac{1}{n})$ with $\frac{1}{n} - ..$ we get $\frac{1}{6^{\alpha}n^{2\alpha}}$. But $\frac{1}{6^{\alpha}}$ plays no role. So what we care about is $\frac{1}{n^{2\alpha}}$. The series of general term $\frac{1}{n^p}$ is called a Riemann series with parameter $p$ or a $p$-series. It is convergent for $p<1$ and divergent otherwise. So convergence $\iff$ $\alpha < \frac{1}{2}$. – user230734 Apr 29 '15 at 17:48