## graydarl 3 years ago I have a sequence that must have lim when n goes to infinity equal to 1 (pic below)

1. graydarl

2. barrycarter

Hint: how many terms are in each element of the sequence, and what are they bounded by?

3. graydarl

each one of them has lim=0?

4. barrycarter

Not quite. There are n terms, each of which is very near 1/n

5. graydarl

it s true, i can see that now, but how can i use this thing?

6. zzr0ck3r

let x_n = 1+1/n

7. graydarl

x_n is just 1+1/n?

8. graydarl

9. ash2326

@graydarl Have you studied sandwich theorem ?

10. ash2326

We have the series as $\lim_{n\to \infty} (\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+....\frac{1}{\sqrt{n^2+n}})$ Obviosuly each of the term is less than 1/n $\frac{1}{\sqrt{n^2+1}}< \frac 1 n$ $\frac{1}{\sqrt{n^2+n}}< \frac 1 n$ so whole series sum will be less than $\frac 1 n + \frac 1 n ....\frac 1 n$ note that there are n terms And if you notice each of the term is greater than 1/(n+1) $\frac{1}{\sqrt{n^2+1}}>\frac{1}{\sqrt{n^2+2n+1}}$ $\frac{1}{\sqrt{n^2+n}}>\frac{1}{\sqrt{n^2+2n+1}}$ so we have $\frac{1}{n+1}+\frac{1}{n+1}....+\frac{1}{n+1}<\frac 1 {\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}....+\frac{1}{\sqrt{n^2+n}}$ $<\frac{1}{n}+\frac{1}{n}.........+\frac{1}{n}$ Now apply the limit $$n\to \infty$$ $\lim_{n\to \infty}(\frac{1}{n+1}+\frac{1}{n+1}....+\frac{1}{n+1})<$ $\lim_{n\to \infty}(\frac 1 {\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}....+\frac{1}{\sqrt{n^2+n}})$ $<\lim_{n\to \infty}(\frac{1}{n}+\frac{1}{n}.........+\frac{1}{n})$ Evaluate the limits for lower and upper bound. If you get the same limit, then by sandwich theorem, you'll be able to prove. Could you try @graydarl

11. ash2326

I mean evaluate these two limits $\Large {\lim_{n\to \infty}(\frac{1}{n+1}+\frac{1}{n+1}....+\frac{1}{n+1})}$ $\Large \lim_{n\to \infty}(\frac{1}{n}+\frac{1}{n}.........+\frac{1}{n})$

12. graydarl

Thank you very much, i understood now, thank you