Help calculate the limits:

1) $$\lim_{n\to \infty}\sum_{k=1}^n \frac{1}{\sqrt {k(n-k)}}$$

2) $$\lim_{n\to \infty}\sum_{k=1}^n \frac{1}{(n-k)\ln{n}} $$

3) $$\lim_{n\to \infty}{n}^p \sin(\pi(\sqrt 2 +1)^n) $$

In the directions to this number it is written: compare $$(\sqrt 2 +1)^n$$ with the whole part $$(\sqrt 2 +1)^n + (-\sqrt 2 +1)^n $$

But I just can not understand how this will help. I got that with even n $$(\sqrt 2 +1)^n + (-\sqrt 2 +1)^n > (\sqrt 2 +1)^n $$ Аnd with odd on the contrary.

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    $\begingroup$ You can try converting the limits to integrals using limit definitions for integrals. $\endgroup$ – Rick May 20 at 18:37
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    $\begingroup$ For the first one, as @Rick suggested you can convert it into an integral and solve it. For the second one, are you sure the question is correct? Because at $k = n$ the expression is undefined. I guess the sum should be till $n-1$ in which case, you can bound the sum above and below using integrals, evaluate them for finite $n$ and then use squeeze theorem. For the third one if $p >0$, then use boundedness of sine along with squeeze theorem to get the limit. These hints should help you get started. $\endgroup$ – sudeep5221 May 20 at 20:11
  • $\begingroup$ @Rick, So it turns out that way? $$\lim_{n\to \infty}\sum_{k=1}^n \frac{1}{\sqrt {k(n-k)}} = \int_1^n \frac{dk}{ \sqrt {k(n-k)}} $$ $\endgroup$ – Mariya May 21 at 19:15
  • $\begingroup$ @sudeep5221, I myself did not understand why in the second example, in the problem book of Makarov, this is exactly what was written $\endgroup$ – Mariya May 21 at 20:05

This suppose to be a comment but it getting kinda long so I placed it here in term of an answer instead, although it's not directly answering your question but it may helps... In regard to Rick's comment, I think this is what he meant:

Suppose I have something like:

$$ \lim_{n \rightarrow \infty} \sum_{k=1}^n \dfrac{1}{\sqrt{k^2 + n^2}} $$

Then I can rewrite this as follow:

$$ \lim_{n \rightarrow \infty} \dfrac{1}{ \sqrt{1^2 + n^2}} + \dfrac{1}{ \sqrt{2^2 + n^2}} + \dfrac{1}{ \sqrt{3^2 + n^2}} + \cdots \dfrac{1}{ \sqrt{n^2 + n^2}} $$

but note that $$ \lim_{n \rightarrow \infty} \dfrac{1}{ \sqrt{1^2 + n^2}} + \dfrac{1}{ \sqrt{2^2 + n^2}} + \cdots + \dfrac{1}{ \sqrt{n^2 + n^2}} = \lim_{n \rightarrow \infty} \dfrac{1}{n} \bigg( \dfrac{1}{\sqrt{ \big(\dfrac{1}{n}\big)^2 + 1 }} + \dfrac{1}{\sqrt{ \big(\dfrac{2}{n}\big)^2 + 1 }} + \cdots + \dfrac{1}{\sqrt{ \big(\dfrac{n}{n}\big)^2 + 1 }} \bigg) $$

From here you can see that it starting to look like how you would define a Riemann sum for the function $ \dfrac{1}{\sqrt{x^2 + 1}} $ from $0$ to $1$ with subdivision point $\dfrac{1}{n}, \dfrac{2}{n}, \cdots $ So now you can evaluate

$$ \lim_{n \rightarrow \infty} \sum_{k=1}^n \dfrac{1}{\sqrt{k^2 + n^2}} = \int_0^1 \dfrac{1}{\sqrt{x^2 + 1}} dx = \log \big[ x + \sqrt(x^2 + 1) \big] \bigg|_0^1 = \log(1 + \sqrt{2} ) $$

For your problem, can you think of doing something similar? Hope this help...


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