4

Using the properties of limits, calculate the following limits, if they exist. If not, prove they do not exist:

$$\lim_{n\to \infty}\left(\frac1{n^2}+\frac1{(n+1)^2}+\frac1{(n+2)^2}+\cdots+\frac1{(2n)^2}\right)$$

This is what I have done, I have expressed the limit in the form:

$\lim_{n\to \infty}\frac1{(n+a)^2}$ where 'a' belongs to the reals.

Then using the $\epsilon-N$ definition of limits, I assumed that:

$$\lim_{n\to \infty}\frac1{(n+a)^2}=0$$ and carried forward with the proof. I would like to use the $\epsilon-N$ definition of limits since it is what we are covering right now, is this the right way of solving this problem?

StubbornAtom
  • 17,052
user323388
  • 155
  • Related : http://math.stackexchange.com/questions/469885/the-limit-of-a-sum-sum-k-1n-fracnn2k2 – lab bhattacharjee Mar 20 '16 at 12:47
  • The convergence to 0 of the individual terms in the summation is, by itself ,insufficient to decide the limit of the sequence of summations because the number of terms in the nth summation goes to infinity. Examples: $\lim_{n\to \infty} \sum {j=n}^{j=2 n}(1/j)=\ln 2$. And $\sum{j=n}^{j=2 n}1/\sqrt n$ goes to infinity as $n$ does. – DanielWainfleet Mar 20 '16 at 13:49

4 Answers4

2

Use these inequalities

$$0\le\frac1{n^2}+\frac1{(n+1)^2}+\frac1{(n+2)^2}+...+\frac1{(2n)^2}\le \frac1n$$ to conclude the desired result.

user296113
  • 7,570
0

Hint $$...=\sum_{k=n}^{2n}\frac{1}{k^2}\leq \frac{n}{n^2}\underset{n\to \infty }{\longrightarrow } 0$$

Surb
  • 55,662
0

Let F(n): $$\frac{1}{n^2}+\frac{1}{(n+1)^2}+\cdots+\frac{1}{(2n)^2}$$ Consider two series: $$G(n): \frac{1}{n^2}+\frac{1}{n^2}+\frac{1}{n^2}+\cdots +\frac{1}{n^2}$$ and $$H(n):\frac{1}{(2n)^2}+\frac{1}{(2n)^2}+\frac{1}{(2n)^2}+\cdots+\frac{1}{(2n)^2}$$ Notice that $$H(n)<F(n)<G(n)$$ Now, $$\lim_{n \to \infty} G(n) = \lim_{n \to \infty} H(n)=0$$ Hence, by the sandwich theorem, the given limit becomes $0$

Nikunj
  • 6,160
  • I understand the point of the squeeze theorem,but if I chose to use the epsilon-n proof, would I be heading in the correct way? – user323388 Mar 20 '16 at 11:50
  • @user323388 I am terribly sorry, I haven't yet covered the epsilon-delta definition of limits, So I can't help with that :( – Nikunj Mar 20 '16 at 11:51
  • @user323388 Where did you use the epsilon-n approach? Your question explains why $1/(n+a)^2$ converges to zero using this method, but not how you would apply it to the sequence of interest. – Did Mar 20 '16 at 12:03
  • @Did I haven't actually applied the epsilon-n proof yet, all I have done is assumed that the limit is 0 and thus using the epsilon-n approach I want to show that the limit is indeed 0 and was unsure whether I was doing it correctly – user323388 Mar 20 '16 at 12:08
0

$\sum_{n\in N}1/n^2$ converges. Therefore $$\lim_{m\to \infty}\sup_{n\geq m} \sum_{j=n}^{j =2 n}(1/n^2)\leq \lim_{m\to \infty}\sup_{n' \geq n\geq m}\sum_{j=n}^{j=n'}(1/n^2)=0.$$

In general if $\sum_{n\in N} a_n$ converges then $\lim_{n\to \infty}\sum _{j=n}^{j=2 n}a_j=0.$

The most elementary way to prove that $\sum_n (1/n^2)$ converges is that for $n>1$ we have $1/n^2<1/n(n-1)=1/(n-1)-1/n$, so for $n>1$ we have $$\sum_{j=1}^n(1/n^2)=1+\sum_{j=2}^n(1/n^2)<1+\sum_{j=2}^n[1/(j-1)-1/j]=2-1/n.$$

DanielWainfleet
  • 57,985