Question

Convergent question.

Answer 1

n^2/(2n-1)*sin(1/n)

Answer 2

The answer is 1/2 but i'm not sure how that was achieved.

Answer 3

also, can someone help with n/2^n thanks!

Answer 4

What are you asking? Are these sequences, and are you asked for the limit?

Answer 5

Sorry. Find the limit of the each convergent sequence.
\[a _{n}=\frac{ n^{2} }{ 2n-1 }(\sin(1/n)\]

Answer 6

Recall that
\[\lim_{x\to0}\frac{\sin x}{x}=\lim_{x\to0}\frac{x}{\sin x}=1\]
Replacing \(x\) with \(\dfrac{1}{x}\), as \(x\to0\) (from the right) you have \(\dfrac{1}{x}\to\infty\), so that
\[\lim_{\frac{1}{x}\to\infty}\frac{\sin \frac{1}{x}}{\frac{1}{x}}=\lim_{x\to0}\frac{\frac{1}{x}}{\sin \frac{1}{x}}=1\]
You have
\[\frac{n^2}{2n-1}\sin\frac{1}{n}=\frac{n}{2n-1}\color{red}{\frac{\sin\frac{1}{n}}{\frac{1}{n}}}\]
and you know that the red term approaches 1 as \(n\to\infty\).

Answer 7

For the second problem, notice that \(\dfrac{n}{2^n}>0\) for all \(n\), but that this sequence is decreasing. To establish this, you have to show that \(a_{n+1}<a_n\), i.e. that
\[\frac{n+1}{2^{n+1}}<\frac{n}{2^n}\]
which you can prove by induction.