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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.

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5 Answers 5

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

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

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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!

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Your answer to part $A$ is correct

For part $B$ use integral test

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

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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 $$\sum_{n=0}^{\infty}q^n=\frac{1}{1-q}\ \text{for all}\ 0<q<1 $$ 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.

See also here:http://home.iitk.ac.in/~psraj/mth101/lecture_notes/Lecture11-13.pdf

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

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