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I have a series $$ \left(\,{\sum_ {k\ge 1} \, {\mid x_k^s \mid}^p \,}\right)^{1 \over p} < \infty $$ where $1 \le p < \infty $, and I want to find the range of values of s in $\mathbb R$ such that $x_k = k$. I solve as below. $$ \left(\,{\sum_ {k\ge 1} \, {\mid x_k^s \mid}^p \,}\right)^{1 \over p} < \infty $$ $$ \implies {\sum_ {k\ge 1} \, {\mid x_k^s \mid}^p \,} < \infty $$ $$ \implies {\sum_ {k\ge 1} \, {x_k^{sp} } \,} < \infty $$ $$ \implies a_n =x_n ^{sp} $$ Using Ratio test $$ L = \lim_{n\rightarrow \infty } \left| {a_{n+1} \over a_n} \right| $$ $$ L = \lim_{n\rightarrow \infty } \left| {x_{n+1}^{sp} \over x_n^{sp}} \right| $$ $$ L = \lim_{n\rightarrow \infty } \left| {(n+1)^{sp} \over n^{sp}} \right|$$ Does the solution correct till this point and how to solve it further to get value of $s \in \mathbb R$

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  • $\begingroup$ That limit is $1$ for all $p,s$ so the ratio test is inconclusive for all $s.$ $\endgroup$ – Thomas Andrews Apr 30 at 0:22
  • $\begingroup$ If you let $r=ps$ what you are really asking is when is $$\sum_{n=1}^{\infty} n^r$$ convergent. Clearly, you need at minimum $r<0,$ because otherwise $n^r$ does not converge to zero. $\endgroup$ – Thomas Andrews Apr 30 at 0:30
  • $\begingroup$ @ThomasAndrews I got the point but I think r<-1 since $$ \sum_{n\ge 1} {1 \over n} $$ is divergent. $\endgroup$ – Math Starter Apr 30 at 0:47
  • $\begingroup$ I didn’t say $r<0$ was sufficient, only that it was necessary. $\endgroup$ – Thomas Andrews Apr 30 at 0:52
  • $\begingroup$ Sure @ThomasAndrews check the answer and give your feedback $\endgroup$ – Math Starter Apr 30 at 0:55
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For $f(x)$ a decreasing positive function on $[1,\infty)$ we have that:

$$f(k) \geq \int_k^{k+1} f(x)\,dx \geq f(k+1)$$ Summing, we get: $$\int_1^{n+1} f(x)\,dx\leq \sum_{k=1}^n f(k)\leq f(1)+\int_1^n f(x)\,dx$$

Taking the limit as $n\to\infty$ on both sides, we see that $\sum_{k=1}^{\infty} f(k)$ converges if and only if $\int_{1}^{\infty} f(x)\,dx$ converges.

Then, if $r<0$ take $f(x)=x^r,$and find out when $$\int_1^\infty x^r\,dx$$ converges.

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$${\sum_ {k\ge 1} \, {x_k^{sp} } \,} < \infty $$ Let r=sp $${\sum_ {k\ge 1} \, {x_k^{r} } \,} < \infty \iff r< -1$$ $$ sp < -1$$ $$ s< {-1 \over p}$$

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