Determine whether the series is convergent or divergent by expressing sn as a telescoping sum (as in example 7). ∞ cos 7 n2 − cos 7 (n + 1)2 n = 1

[tex]\displaystyle\sum_{n=1}^k\cos\frac7{n^2}-\cos\frac7{(n+1)^2}=[/tex][tex]\left(\cos7-\cos\dfrac7{2^2}\right)+\left(\cos\dfrac7{2^2}-\cos\dfrac7{3^3}\right)+\cdots+\left(\cos\dfrac7{k^2}-\cos\dfrac7{(k+1)^2}\right)[/tex]
We can see some cancellation occurring that leaves us with
[tex]\displaystyle\sum_{n=1}^k\cos\frac7{n^2}-\cos\frac7{(n+1)^2}=\cos7-\cos\frac7{(k+1)^2}[/tex]
Then as [tex]k\to\infty[/tex], we have [tex]\dfrac7{(k+1)^2}\to0[/tex] and [tex]\cos0=1[/tex], so the series converges to [tex]\cos7-1[/tex].

MrRoyal MrRoyal · Answer 2 · 2022-03-24T14:52:07+01:00

The series converges to [tex]\cos 7 - 1[/tex]

The series is given as:

[tex]\sum\limits^{\infty}_{n=1}\cos \frac{7}{n^2} - \cos \frac{7}{(n + 1)^2}[/tex]

Assume the series is to k terms.

So, we have:

[tex]\sum\limits^{k}_{n=1}\cos \frac{7}{n^2} - \cos \frac{7}{(n + 1)^2} = (\cos \ 7 - \cos \frac{7}{2^2}) + (\cos \ \frac{7}{2^2} - \cos \frac{7}{3^2}) + .... + (\cos \ \frac{7}{k^2} - \cos \frac{7}{(k + 1)^2})[/tex]

Expand

[tex]\sum\limits^{k}_{n=1}\cos \frac{7}{n^2} - \cos \frac{7}{(n + 1)^2} = \cos \ 7 - \cos \frac{7}{2^2}+ \cos \ \frac{7}{2^2} - \cos \frac{7}{3^2}+ .... + \cos \ \frac{7}{k^2} - \cos \frac{7}{(k + 1)^2}[/tex]

Cancel out the common terms

[tex]\sum\limits^{k}_{n=1}\cos \frac{7}{n^2} - \cos \frac{7}{(n + 1)^2} = \cos \ 7 - \cos \frac{7}{(k + 1)^2}[/tex]

The above means that:

As k approaches infinity, [tex]\cos \frac{7}{(k + 1)^2}[/tex] approaches 0

So, we have:

[tex]k \to \infty[/tex], [tex]\frac{7}{(k + 1)^2} \to 0[/tex]

So, we have:

[tex]\cos(0) = 1[/tex]

So, we have:

[tex]\sum\limits^{\infty}_{n=1}\cos \frac{7}{n^2} - \cos \frac{7}{(n + 1)^2} \to \cos 7 - 1[/tex]

Hence, the series converges to [tex]\cos 7 - 1[/tex]