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I would like to prove $$\sum\limits_{k=n}^{\infty}\frac{1}{k^2}\leq \frac{C}{n},$$where $C$ is independent of $n$.

My Attempt:

Since $\sum\limits_{k=n}^{\infty}\frac{1}{k^2}$ is convergent, there exists $N=N(n)$ such that if $k>N$, there holds $$\sum\limits_{k=N+1}^{\infty}\frac{1}{k^2}<\frac1n$$ This implies that $$\sum\limits_{k=n}^{\infty}\frac{1}{k^2}= \sum\limits_{k=n}^{N}\frac{1}{k^2}+\sum\limits_{k=N+1}^{\infty}\frac{1}{k^2}<\frac{N-n+1}{n^2}+\frac1n$$ However, I found the integer $N$ here is dependent of $n$.

Is there a smart and correct way to prove this inequality? Any insight or help is appreciated.

MathFail
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Louise Wing
  • 23

1 Answers1

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$$\frac1n=\int_{n}^\infty\frac1{x^2}dx\le\sum_{k=n}^{\infty}\frac{1}{k^2}\le \int_{n-1}^\infty\frac1{x^2}dx=\frac1{n-1}$$

MathFail
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