# What test should I use to prove the convergence of the following series?

Prove the convergence or divergence of the following series

A) $$\sum_{n=1}^\infty \frac{1}{2^n +n}$$

B) $$\sum_{n=1}^\infty \frac{ln(n)}{n}$$

C) $$\sum_{n=1}^\infty tan(\frac{1}{n \sqrt (n)})$$

In A) I tried using the direct comparison test: I wrote $$2^n +n>2^n$$ so $$\frac{8}{2^n +n} < \frac{8}{2^n}$$. And as $$8. \frac{1}{2^n}$$ converges, the other one converges. Is it ok?

In B) I also thought about direct comparison test, but I don't know what series I should use to compare.

In C) I don't know what test to use.

For B use $$\frac{\ln{(n)}}{n}\gt\frac1n$$ Then for C use $$\tan\left(\frac1{n^{3/2}}\right)\lt\frac2{n^{3/2}}$$

• But $\frac{ln(n)}{n}$ is not always bigger than $\frac{1}{n}$ does it matter? Aug 4, 2019 at 22:23
• As long as it is true for infinitely many $n$ (above a certain number) then the comparison test works. Aug 4, 2019 at 22:25

Your method for A) is good. You don't have to use $$\frac{8}{2^n}$$, you can just use $$\frac{1}{2^n}$$

For B), you can use direct comparison with $$\sum_{n=1}^{\infty}\frac{1}{n}$$

For C), you can use direct comparison with $$\sum_{n=1}^{\infty}\frac{2}{n\sqrt{n}}$$ which is then a p-series.

First one is correct!

For second one, we have $$\ln(x) > 1$$ after $$x = e$$, so we can start the sum from $$n>e$$ or $$n \geq 3$$.So, $$\sum_{n=3}^{\infty} \frac{\ln(n)}{n} > \sum_{n=3}^{\infty} \frac{1}{n}$$, so by limit comparision test the series diverges!, also you can work out the same using integral test, approximating the sum as the integral as $$\sum_{n=1}^{\infty} \frac{\ln(n)}{n} \approx \int_{1}^{\infty} \frac{\ln(x)}{x}dx$$.

For the third we have for large $$n$$, $$\tan(\frac{1}{n\sqrt{n}})$$ is smaller and we can use the approximation $$\tan(x) \approx x$$ for small $$x$$. After this you can use the limit comparision test to show the convergence of the sum of the series!

Your answer to part $$A$$ is correct

For part $$B$$ use integral test

For part $$C$$ use limit comparison test with $$\sum _1^{\infty} \frac {1}{n\sqrt n}$$

Note first that a necessary condition for convergence of a series $$\sum_{n=1}^{\infty}a_{n}$$ is that the sequence $$a_{n}$$ tends to zero for $$n\to \infty$$.

Once you observe that $$|a_{n}|$$ doesn't become arbitrarily small as n increases you already know that the series must diverge.

In A,B the necessary condition for convergence is satisfied

In A the comparison $$\frac{1}{2^n+n}<\frac{1}{2^n}$$ works just fine since we know that the series $$\sum_{n=1}^{\infty}\frac{1}{2^n}$$ converges to 1. In case you might not know or remember,the last fact is due to the following fact about geometric series $$\sum_{n=0}^{\infty}q^n=\frac{1}{1-q}\ \text{for all}\ 0 and follows from this formula we can derive for the partial sums $$\sum_{k=0}^{n}q^k=\frac{1-q^{n+1}}{1-q}$$.

For part B note that the comparison test works in two directions

If $$0\leq a_{n}\leq b_{n}$$ for $$n\geq k$$ for some k ,then

1) Convergence of $$\sum_{n=1}^{\infty}b_{n}$$ implies convergence of $$\sum_{n=1}^{\infty}a_{n}$$

2) Divergence of $$\sum_{n=1}^{\infty}a_{n}$$ implies divergence of $$\sum_{n=1}^{\infty}b_{n}$$

And then consider the harmonic series $$\sum_{n=1}^{\infty}\frac{1}{n}$$,

which is a series which satisfies the necessary condition for convergence but is divergent.

If k is choosen sufficiently large $$\frac{1}{n}\leq \frac{ln(n)}{n}$$ for all $$n\geq k$$

This implies that the series $$\sum_{n=1}^{\infty}\frac{ln(n)}{n}$$ must diverge.

https://en.wikipedia.org/wiki/Harmonic_series_(mathematics)