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I need to check if the series is convergent:

$$\sum_{l=2}^\infty{\frac{1}{l^m}}, m \ge 3$$

On the one hand I think that it has to be divergent because each iteration there will be added something more like the harmonic series.

On the other hand I think it has to be convergent because to some extend the sum value will be so small that it does not matter anymore.

Martin Sleziak
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jublikon
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2 Answers2

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Consider the grouping

$$\frac1{2^m}+\frac1{3^m}+\\ \frac1{4^m}+\frac1{5^m}+\frac1{6^m}+\frac1{7^m}+\\ \frac1{8^m}+\frac1{9^m}+\frac1{10^m}+\frac1{11^m}+\frac1{12^m}+\frac1{13^m}+\frac1{14^m}+\frac1{15^m}+\\\cdots $$ with doublings of the number of terms.

As every term in a group is bounded above by its first element $1/2^{km}$, and as there are $2^k$ terms per group, a group contributes less than $2^{(1-m)k}$, forming a geometric series of common ratio $2^{1-m}$.

$$S<\frac2{2^m}+\frac4{4^m}+\frac8{8^m}+\cdots$$

You can conclude about convergence.

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Setting $$ S_N:=\sum_{l=2}^N{\frac{1}{l^m}}, \quad m \ge 3 $$ we have $$ S_{N+1}-S_N=\frac{1}{l^m}>0, $$ $\{S_N\}$ is an increasing sequence then observing that, for $m\ge3$, $$ S_N=\sum_{l=2}^N{\frac{1}{l^m}}<\sum_{l=2}^N{\frac{1}{l(l+1)}}=\frac12-\frac1{N+1}<\frac12 $$ $\{S_N\}$ is then a bounded sequence, the given series is thus convergent.

Olivier Oloa
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    can the same be applicable if $m>1$ ? your ans is beautiful – KON3 Apr 26 '20 at 03:41
  • don't you think it is $\frac{1}{l(l-1)}$ instead of $\frac{1}{l(l+1)}$? – KON3 Apr 26 '20 at 05:58
  • @Anjan3 Thank you. Since $m\ge3$ and $l\geq2$, then it's OK that $\frac{1}{l^m}<\frac{1}{l(l+1)},$ as one may check. A proof for $m>1$ probably needs to compare $\frac{1}{l^m}$ with $\frac{1}{l(l-1)}$ as you suggest. – Olivier Oloa Apr 26 '20 at 09:52