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Yesterday at university a professor gave us two problems that left many doubts.

1) $\displaystyle \sum_{i=1}^n \frac{1}{i^2} \leq 2-\frac1{n}$,

2) $\displaystyle \sum_{i=1}^n \frac1{n+i} \leq \frac3{4}$.

I tried to solve both using the "classic" induction way initially, so I try for $n=0$ first, then I assume $n$ is right and finally I try with $n+1$; with some shift in the summation etc, in both situation I could say that the second term of "$n+1$" is always bigger than the second term of "$n$", so if I say so, is this enough? Are there other way to solve this? Thanks a lot!

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As regards the first sum, we have that $$\sum_{i=1}^n \frac{1}{i^2}\leq 1+\sum_{i=2}^n \frac{1}{i(i-1)}= 1+\sum_{i=2}^n \left(\frac{1}{i-1}-\frac{1}{i}\right)=1+1-\frac{1}{n}=2-\frac{1}{n}.$$

Hint for the second sum. Show that for $i=1,\dots,n$, $$\frac1{n+i}+\frac1{n+(n+1-i)}\leq \frac{3}{2n}.$$

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  • $\begingroup$ So sorry, I edited the second and now it should be right $\endgroup$ – L. Repetti Oct 18 '16 at 14:30
  • $\begingroup$ Sorry for the last comment, I failed to use stackechange, I am new to this site; anyway, for the first sum, how do you "make" i(i-1)? And for the second sum, I really don't understand how you did that $\endgroup$ – L. Repetti Oct 18 '16 at 14:46
  • $\begingroup$ @L. Repetti The sum of $1/(i(i-1)$ is "easier" than $1/i^2$. In 2), take the sum of the symmetric terms. Anyway take a look to the related questions on the right. There are similar helpful exercises. Then see also: math.stackexchange.com/tour $\endgroup$ – Robert Z Oct 18 '16 at 14:56
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For the first, one can establish

$$S_{n+1}=S_n+\frac1{(n+1)^2}\le 2-\frac1n+\frac1{(n+1)^2}\le2-\frac1{n+1},$$

which is true because $-(n+1)^2+n\le-n(n+1)$.

For the second,

$$S_{n+1}=S_n-\frac1{n+1}+\frac1{2n+1}\le\frac34-\frac1{n+1}+\frac1{2n+1}\le\frac34$$ because $n+1<2n+1$.

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