Show that the sequence $y_n = 1+\frac{1}{2^k}+\frac{1}{3^k}+...+\frac{1}{n^k}$, where $n\in \mathbb{N}$ and $k\geq 2$ is convergent.

Asked Oct 21 '18 at 09:13

Active Oct 21 '18 at 09:29

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Let

$y_n = 1+\frac{1}{2^k}+\frac{1}{3^k}+...+\frac{1}{n^k}$, where $n\in \mathbb{N}$ and $k\geq 2$.

I need to show that this sequence is convergent.

I did $y_{n+1}-y_n$ and I got that it is bigger than $0$ so the sequence is growing and I know that $y_1 \leq y_n$ but I need to also find a number which is bigger then $y_n$ which I do not know how to do. The answers says that I need to show that the sequence is smaller than $2$ but I really do not have any idea how to do it.

edited Oct 21 '18 at 09:29

asked Oct 21 '18 at 09:13

Ghost

1,105

To be fair this was quite hard to find. – Kenny Lau Oct 21 '18 at 09:20

Show that the sequence $y_n = 1+\frac{1}{2^k}+\frac{1}{3^k}+...+\frac{1}{n^k}$, where $n\in \mathbb{N}$ and $k\geq 2$ is convergent.

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