Question

determine whether the series is convergent or divergent by expressing the nth partial sum as a telescoping series. If it converges find the limit.

(equation given inside)

Answer 1

\[\sum_{n=1}^{\infty} (\cos1/n^2 - \cos1/(n+1)^2\]

Answer 2

If
\[S=\sum_{n=1}^\infty\left(\cos\frac{1}{n^2}-\cos\frac{1}{(n+1)^2}\right)\]
then the \(N\)th partial sum is
\[\begin{align*}S_N&=\sum_{n=1}^N\left(\cos\frac{1}{n^2}-\cos\frac{1}{(n+1)^2}\right)\\\\
&=\left(\cos1\color{red}{-\cos\frac{1}{4}}\right)+\left(\color{red}{\cos\frac{1}{4}}\color{blue}{-\cos\frac{1}{9}}\right)+\left(\color{blue}{\cos\frac{1}{9}}-\cos\frac{1}{16}\right)\\
&\quad\quad+\cdots+\left(\cos\frac{1}{N^2}-\cos\frac{1}{(N+1)^2}\right)\end{align*}\]
From the pattern above, which terms do you think disappear, and what remains?

Answer 3

so you'll end up with cos1+cos1/N^2 -cos1/(N+1)^2?

Answer 4

\[\cos1+\cos1/N^2 - \cos1/(N+1)^2\]

Answer 5

Almost, you have an extra term there. From the color matching, you know that the negative terms will cancel out so long as the positive term is also there. You know that \(-\cos\dfrac{1}{16}\) because the next term will contain \(+\cos\dfrac{1}{16}\). For the same reason, \(\cos\dfrac{1}{N^2}\) will also cancel, so you'd be left with just \(\cos1-\cos\dfrac{1}{(N+1)^2}\).

Answer 6

Oh okay I see!

Answer 7

what would I do next?

Answer 8

You have the \(N\)th partial sum now,
\[S_N=\cos1-\cos\frac{1}{(N+1)^2}\]
which is equivalent to the infinite sum if you take the limit:
\[S=\lim_{N\to\infty}S_N=\lim_{N\to\infty}\left(\cos1-\cos\frac{1}{(N+1)^2}\right)=\cdots\]
If the limit exists and is finite, you know the sum must converge.

Answer 9

It is convergent. So would the limit be -1/2?

Answer 10

Hmm, not quite... The first term is independent of \(N\), so you have
\[S=\lim_{N\to\infty}\left(\cos1-\cos\frac{1}{(N+1)^2}\right)=\cos1-\lim_{N\to\infty}\cos\frac{1}{(N+1)^2}\]
The cosine function is continuous for all \(N\), so you have
\[S=\cos1-\cos\left(\lim_{N\to\infty}\frac{1}{(N+1)^2}\right)\]
What's the value of this limit?

Answer 11

0

Answer 12

Right, so
\[S=\cos1-\cos0=\cdots\]

Answer 13

cos1-1

Answer 14

Right, that's the answer then

Answer 15

perfect thank you again so much youre VERY helpful!

Answer 16

You're welcome!