Question

I have a sequence that must have lim when n goes to infinity equal to 1 (pic below)

Answer 1

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

Answer 2

each one of them has lim=0?

Answer 3

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

Answer 4

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

Answer 5

let x_n = 1+1/n

Answer 6

x_n is just 1+1/n?

Answer 7

Please explain further

Answer 8

@graydarl Have you studied sandwich theorem ?

Answer 9

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

Answer 10

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



\[\Large \lim_{n\to \infty}(\frac{1}{n}+\frac{1}{n}.........+\frac{1}{n})\]

Answer 11

Thank you very much, i understood now, thank you