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  1. $$ \lim_{n \to \infty}( \frac {n}{(n+1)^2} +\frac {n}{(n+2)^2} + ... +\frac {n}{(2n)^2 }) $$
  2. $$ \lim_{n \to \infty} \frac {1^{\alpha} + 2^{\alpha} + ... + n^{\alpha}}{n^{\alpha +1}}$$
  3. $$\lim_{n \to \infty} \frac {(n!)^{\frac {1}{n}}} {n} $$

(can probably be handeled without this trick, but I am wondering how to use it here) I know the main idea -- the sums above is nothing but Riemann sums for some function and some partition (and actually doesn't depend on partition). So, I gotta see needed function and partition and use integral to calculate limit. But I can't notice function and even segment.

Tianlalu
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Invincible
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  • for the first we get $$\frac{1}{2}$$ as the searched limit – Dr. Sonnhard Graubner Dec 18 '16 at 10:30
  • I need the function and segment $[a,b]$ more than value of limit. – Invincible Dec 18 '16 at 10:32
  • Use http://math.stackexchange.com/questions/469885/the-limit-of-a-sum-sum-k-1n-fracnn2k2 for the first two and http://math.stackexchange.com/questions/475786/how-to-compute-lim-n-rightarrow-infty-frac1n-left-2n12n2-cdots2nn for the last . But please put one question at a time with your effort. – lab bhattacharjee Dec 18 '16 at 10:33

3 Answers3

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Hint. As regards 1., note that $$ \frac {n}{(n+1)^2} +\frac {n}{(n+2)^2} + ... +\frac {n}{(2n)^2 }=\frac{1}{n}\sum_{k=1}^n\frac{1}{\left(1+\frac{k}{n}\right)^2}.$$ What is the corresponding integral?

For 2. try using a similar approach.

For 3. take the logarithm and evaluate $$\ln\left(\frac {(n!)^{\frac {1}{n}}} {n}\right)=\frac{1}{n}\sum_{k=1}^n\ln(k)-\ln(n).$$ Can you take it from here?

Robert Z
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  • What is the corresponding integral?

    I am probably stupid, but I didn't get even what is corresponding segment.

     – Invincible Dec 18 '16 at 10:44
  • @Vladislav The points of the partition are $k/n$ for $k=1,\dots,n$. So what is the segment? For the function replace $k/n$ with $x$. – Robert Z Dec 18 '16 at 10:47
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Hint (3)

$$\log \frac{(n!)^{1/n}}{n} = \frac{1}{n}\log \prod_{k=1}^n \frac{k}{n} = \frac{1}{n} \sum_{k=1}^n \log \frac{k}{n}$$

RRL
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  • It is not true: $\sum_{k=1}^n \log \frac{k}{n} = log( \frac {n!} {n^n}$ – Invincible Dec 18 '16 at 12:04
  • I am suggesting the same since $\log \frac{(n!)^{1/n}}{n} = \log \left( \frac{n!}{n^n}\right)^{1/n} = \frac{1}{n} \log \frac{n!}{n^n} = \frac{1}{n} \log \prod_{k=1}^n \frac{k}{n} = \ldots$ – RRL Dec 18 '16 at 20:18
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Hint for 2: $$\lim _{ n\to \infty } \frac { 1^{ \alpha }+2^{ \alpha }+...+n^{ \alpha } }{ n^{ \alpha +1 } } =\lim _{ n\to \infty } \frac { 1 }{ n } \left( \frac { 1^{ \alpha } }{ n^{ \alpha } } +\frac { 2^{ \alpha } }{ n^{ \alpha } } +...+\frac { n^{ \alpha } }{ n^{ \alpha } } \right) =\lim _{ n\to \infty } \frac { 1 }{ n } \sum _{ k=1 }^{ n }{ \frac { { k }^{ \alpha } }{ { n }^{ \alpha } } } $$

haqnatural
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