# Convergence of series and finding range of other variable.

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 $$s \in \mathbb R$$

• That limit is $1$ for all $p,s$ so the ratio test is inconclusive for all $s.$ – Thomas Andrews Apr 30 at 0:22
• 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. – Thomas Andrews Apr 30 at 0:30
• @ThomasAndrews I got the point but I think r<-1 since $$\sum_{n\ge 1} {1 \over n}$$ is divergent. – Math Starter Apr 30 at 0:47
• I didn’t say $r<0$ was sufficient, only that it was necessary. – Thomas Andrews Apr 30 at 0:52
• Sure @ThomasAndrews check the answer and give your feedback – Math Starter Apr 30 at 0:55

## 2 Answers

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.

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