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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 $\text{p < 1}$.

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    $\begingroup$ 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. $\endgroup$ – Mark Viola Feb 25 '16 at 4:04
  • $\begingroup$ @Dr.MV edit and corrected. $\endgroup$ – vivek Feb 25 '16 at 4:09
  • $\begingroup$ And now, you've answered your own question. $\endgroup$ – Mark Viola Feb 25 '16 at 4:10
  • $\begingroup$ @Dr.MV thanks for correcting me. $\endgroup$ – vivek Feb 25 '16 at 4:11
  • $\begingroup$ You're welcome. My pleasure! - Mark $\endgroup$ – Mark Viola Feb 25 '16 at 4:15
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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 $\frac{\pi^2}{6}$.

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  • $\begingroup$ You have applied the comparison test right? $\endgroup$ – vivek Feb 25 '16 at 4:10
  • $\begingroup$ @vivek Yes, I think so.. I don't know what it is called. $\endgroup$ – S.C.B. Feb 25 '16 at 4:10

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