$\sum_{n=1}^\infty 1/n^{1+\epsilon}$ Converges for all $\epsilon >0$

Question

If $\epsilon>0$, Prove that the sum$$\sum_{n=1}^\infty \frac{1}{n^{1+\epsilon}}$$ converges.

This problem is from my calculus book, and Since this problem follows some other problems like: Show that $\sum_{n=1}^\infty 1/n^2$ And $\sum_{n=1}^\infty 1/n^3$ converges, i tried to use the same method that i used to solve these problems, but it didn’t work.

The method is very simple, all what you have to do is to find a series that you know that it definitely converges and you also know that this series Is less or equal to your series.

The series i used to prove that the above series converges is $$\sum_{n=0}^\infty \frac{1}{2^n}$$ But again it didn’t help me in this $\sum_{n=1}^\infty \frac{1}{n^{1+\epsilon}}$

Are you known to $p$-series? Any series of the form $\sum_{n=1}^\infty 1/n^p$ converges for $p>1$ and diverges for $p\le1$. But is there in your question any typo? There's nothing $i$ in the summand. — Manjoy Das, Mar 24 '21 at 17:30
Go through this please. https://mathcs.clarku.edu/~djoyce/ma122/posseries.pdf — Manjoy Das, Mar 24 '21 at 17:32
@Yassir If my answer has helped you, I'd appreciate a checkmark. — vitamin d, Mar 24 '21 at 18:43
Well i will do, but first answer my question in the comments. @vitamind — PNT, Mar 24 '21 at 18:47

score 2 · Accepted Answer · answered Mar 24 '21 at 17:50

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$$\sum_{n=1}^\infty \frac{1}{n^{1+\epsilon}}\leq\int_{1}^\infty \frac{1}{n^{1+\epsilon}}\,\mathrm{d}n=\frac{1}{\varepsilon}-\lim_{n\to\infty}\frac{n^{-\varepsilon}}{\varepsilon}$$ Only converges for $\varepsilon>0$.

answered Mar 24 '21 at 17:50

vitamin d

5,783

Is the integral always less than the series? – PNT Mar 24 '21 at 18:44
@Yassir Here is the convergence test. Very useful in many different cases! – vitamin d Mar 24 '21 at 18:50

score 0 · Answer 2 · answered Mar 24 '21 at 17:38

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Since this is a decreasing positive sequence, the Cauchy Condensation test is applicable. The condensed series becomes $\displaystyle\sum_{n=1}^\infty 2^{-n\epsilon}$ in this case, which is a GP with common ratio less than $1$.

answered Mar 24 '21 at 17:38

Martund

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score -2 · Answer 3 · answered Mar 24 '21 at 17:43

In this case you should use the p-series test. The p-series test states that the infinite series $$\sum_{n=1}^\infty \frac{1}{n^p}$$ converges only if p > 1. For p < 1 or p = 1 (harmonic series), this series diverges. In your case, you have p = $\epsilon$ + 1. From there it's pretty straight-forward. Because $\epsilon$ by definition has to be greater than zero, it insures that p must be greater than 1. The p=series test therefore confirms that the series converges.

You are pretty much assuming the result you are trying to prove. — Blitzer, Mar 24 '21 at 18:44

$\sum_{n=1}^\infty 1/n^{1+\epsilon}$ Converges for all $\epsilon >0$

3 Answers3