3
$\begingroup$

I don't really know how the Squeeze theorem works and I tried applying it to solve this limit:

$$\lim_{n\to\infty}\text{ } \frac{1}{n^2}+\frac{1}{(n+1)^2}+\dots+\frac{1}{(n+n)^2}$$ So $$\frac{n}{(n+n)^2}\leq \frac{1}{n^2}+\frac{1}{(n+1)^2}+\dots+\frac{1}{(n+n)^2} \leq \frac{n}{n^2} $$ Then:

$$\lim_{n\to\infty}\text{ } \frac{n}{n^2}=0$$

$$\lim_{n\to\infty}\text{ } \frac{n}{(n+n)^2}=0$$

Therefore: $$\lim_{n\to\infty}\text{ } \frac{1}{n^2}+\frac{1}{(n+1)^2}+\dots+\frac{1}{(n+n)^2}=0$$

Is this the correct way?

$\endgroup$
2
$\begingroup$

You should have $n \to \infty$ not $x \to \infty$.

You can afford to be sloppier with the lower bound ($0$ will work), and your upper bound will be much easier to verify if you use $\dfrac{n+1}{n^2}$.

$\endgroup$
2
$\begingroup$

what you have is: $$L=\lim_{n\to\infty}\sum_{k=0}^n\frac{1}{(n+k)^2}$$ and if we make the substitution $u=n+k$ we can obtain: $$L=\lim_{n\to\infty}\sum_{u=n}^{2n}\frac{1}{u}^2$$ which we can rewrite as: $$L=\lim_{n\to\infty}\left[\sum_{u=0}^{2n}\frac{1}{u^2}-\sum_{u=0}^{n-1}\frac{1}{u^2}\right]$$ and when $n\to\infty$ these two summations are equal, so it is equal to $0$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.