Question

my final exam (calculus 1) 5 years ago

\[\lim_{n \rightarrow \infty} \sum_{k=1}^{\infty} (\sqrt{1+\frac{k}{n^2}}-1)\]

Answer 1

\(\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\pi}}}}}}}}}}\)

Answer 2

What's your question?

Answer 3

Congratulations For Golden benim yoldashim..

@mukushla

Answer 4

tnx mere yaar

Answer 5

Nice problem. But I cant find the solution. I will think about and try to solve it.

Answer 6

I don't like that \(k\) has 1 degree in radicand and \(n\) has 2.

Answer 7

yes nice one ...worth to try...

Answer 8

I tried to use \[\lim_{n\rightarrow \infty}\left( \frac{b-a}{n}\cdot\sum_{k=1}^{n} \ f\left(a+k\cdot\frac{b-a}{n} \right)\right)=\int_a^bf(x)\ dx\]If \(f\) is defined on \([a,b]\). \(a=0,\ b=1\). Rewrite the limit in other form:
\[\lim_{n\rightarrow \infty}\left( \frac{1}{n}\cdot\sum_{k=1}^{n} \ n\cdot \left(\sqrt{1+\frac1 n \cdot\frac{k}{n}}-1\right)\right)\]Calculate the integral:
\[\int_0^1\ n\cdot \left(\sqrt{1+\frac x n }-1\right)\ dx=-n + \frac23 \left(-1 + \left(1 + \frac 1n\right)^{\frac32}\right) n^2\]If we calculate the limit we will get:
\[\lim_{n \rightarrow \infty} \sum_{k=1}^{\infty} \left(\sqrt{1+\frac{k}{n^2}}-1\right)=0\]

Answer 9

But I cant understand why zero.

Answer 10

I tried to calculate it numerically with Mathematica. I put k and n = 1000000 fnd it seems to be 0, but i cant understand how it can be.

Answer 11

ahh... i thought it has a value
lets think again...

Answer 12

lol .. that would make double sum!! for n = k = 10000 mathematica seems to take forever. anyone doesn't own a supercomputer?? lol

Answer 13

Did anybody understand my solution?

Answer 14

yes i workin on it...

Answer 15

lol ... we aren't summing n !!

Answer 16

sorry guys i lost my connection

@klimenkov
no problem with ur approach u just calculated lim wrong

\[\lim_{n \rightarrow \infty} \left( \frac{2}{3}n^2 \left( (1+\frac{1}{n})^{\frac{3}{2}}-1 \right)-n\right)=\frac{1}{4}\]

Answer 17

We have to divide it by n and finally get zero.

Answer 18

here is my solution

Using Taylor series
\[ (1+\frac{k}{n^2})^{\frac{1}{2}} \approx1+\frac{k}{2n^2} \]

then


\[\lim_{n \rightarrow \infty} \sum_{k=1}^{\infty} (\sqrt{1+\frac{k}{n^2}}-1)=\lim_{n \rightarrow \infty} \sum_{k=1}^{\infty} (1+\frac{k}{2n^2}-1)=\lim_{n \rightarrow \infty} \sum_{k=1}^{\infty} \frac{k}{2n^2}\\=\lim_{n \rightarrow \infty} \frac{n(n+1)}{4n^2}=\frac{1}{4}\]