How to check if given series converges?

Question

Given this series:

$$\sum_{m=5}^\infty\left(\frac 7{5m^{5.8}} \right)$$

I can write is as:

$$\frac 75 \sum_{m=5}^\infty\left(\frac 1{m^{5.8}} \right)$$

Now, it seems $\frac {1}{m^{5.8}}$ is a Harmonic series hence it should diverge. How to proceed from here?

$$$$ Edit

The above series is a $P$ series i.e. series of the form:

$$ \sum_{n=1}^\infty \frac 1{n^p} $$

which means the series converges if $\text{p > 1}$ and diverges if $\text{p < 1}$.

It is not harmonic. The harmonic series is $\sum_{k=1}^{\infty} \frac1k$, which diverges. Now, for any number $\alpha>1$, the series $\sum_{k=1}^{\infty} \frac1{k^\alpha}$ converges. For the series of interest, $\alpha =5.8>1$. Therefore, the series of interest converges. — Mark Viola, Feb 25 '16 at 04:04

score 1 · Accepted Answer · edited Apr 13 '17 at 12:20

1

Note that the sequences is not convergent. From Basel's Problem, notice that $$\sum_{k=1}^\infty \frac{1}{k^{5.8}} \le \sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$$

For the proof, see here.

This implies that $$\sum_{k=5}^\infty \frac{1}{k^{5.8}} \le \sum_{k=1}^\infty \frac{1}{k^{5.8}} \le \frac{\pi^2}{6}$$

Therefore, your series is convergent, and converges to a value smaller than $\frac{\pi^2}{6}$.

edited Apr 13 '17 at 12:20

Community

1

answered Feb 25 '16 at 04:05

S.C.B.

22,768

You have applied the comparison test right? – vivek Feb 25 '16 at 04:10
@vivek Yes, I think so.. I don't know what it is called. – S.C.B. Feb 25 '16 at 04:10

How to check if given series converges?

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