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The series in question is: $$\sum_{n=1}^{\infty }n^2\left(\cos\left(\frac{1}{n}\right)-1+\frac{1}{2n^2}\right)$$ and I need to see whether or not it converges.

What I tried was taking the absolute of the term, and finding its limit (to check whether it diverges by $n-$th term test). I obtained: $$\begin{align} \lim_{n\to \infty }n^2\left|\cos \left(\frac{1}{n}\right)-1+\frac{1}{2n^2}\right| &= \lim_{n\to \infty }n^2 \left|1-1+\frac{1}{2n^2}\right| \\ &= \lim_{n\to \infty }n^2 \left(\frac{1}{2n^2}\right)\\ &=\frac{1}{2}. \end{align}$$ The series is supposed to diverge as the limit of its partial sums is different from zero. I think the answer is supposed to be that the series converges according to graphing software, but I am unable to spot the error in my calculations.

M. A.
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user133287
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    A very common mistake. You can't compute the limit by parts, changing $\cos(\frac{1}{n})$ to $1$ while keeping the other terms with $n$. This is just false. The limit of the general term is equal to $0$, this can be easily checked using L'hopital's rule. – Mark Oct 31 '24 at 19:57
  • I see now, thanks! In that case, how would one go on about solving a series such as this when when trig functions are involved? – user133287 Oct 31 '24 at 20:01
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    use the Taylor expansion of the cosine – Sine of the Time Oct 31 '24 at 20:01
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    @user133287 The terms after $\cos$ should remind you of the first few terms of the Taylor expansion of $\cos(x)$. Write down the expansion and see what you get if you substitute $x=\frac{1}{n}$. – Mark Oct 31 '24 at 20:04
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    You first just let one $n$ go to infinity, then the other? By this reasoning $$\lim_{n}n\cdot\frac1n=\lim_{n}n\cdot 0=0.$$ – Thomas Andrews Oct 31 '24 at 20:09
  • @Mark Viola showed the convergence. If you continue the expansion, you should get the "exact" value. of the summation. – Claude Leibovici Nov 01 '24 at 10:09
  • In fact, you do not need to know anything at all about Taylor's Theorem or Taylor's Series. Please see my posted solution for a way forward that relies on pre-calculus tools only – Mark Viola Nov 01 '24 at 19:57
  • Please let me know how I can improve my answer. I really want to give you the best answer I can. – Mark Viola Dec 07 '24 at 05:10
  • @user133287 And please feel free to up vote and accept an answer as you see fit. ;-) – Mark Viola Dec 07 '24 at 05:10

2 Answers2

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I thought it might be instructive to post a solution that does not rely on use of Taylor's theorem, but instead uses trigonometric relations and inequalities only. To that end, we now proceed.

Note that we have

$$1-\cos\left(\frac1n\right)=2\sin^2\left(\frac{1}{2n}\right)\tag1$$

Furthermore, in THIS ANSWER, I showed using basic trigonometry only that the sin function satifies the inequalities for $x>0$

$$x-\frac{x^3}{6}<\sin(x)< x\tag2$$

Using $(1)$, we can write for $n\ge 1$

$$\begin{align} n^2\left(\cos\left(\frac1n\right)-1+\frac1{2n^2}\right)&=-n^2\left(2\sin^2\left(\frac{1}{2n}\right)-\frac1{2n^2}\right)\\\\ \end{align}$$

Then, using $(2)$, we find that for $n\ge 1$

$$0<-n^2\left(2\sin^2\left(\frac{1}{2n}\right)-\frac1{2n^2}\right)<\frac1{24n^2}-\frac1{48n^4}$$

Inasmuch as $\sum_{n=1}^\infty \frac1{n^2}$ converges, the series of interest does likewise. And we are done!

Mark Viola
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As suggested by comments above, we start with Taylor expansion of cosine (around zero): $$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + ...$$ Thus, $$\cos x -1 < - \frac{x^2}{2!} + \frac{x^4}{4!} $$

Substituting this in the summation term: $$n^2\left(\cos\left(\frac{1}{n}\right)-1+\frac{1}{2n^2}\right) < n^2\left(-\frac{1}{2n^2}+\frac{1}{4!n^4}+\frac{1}{2n^2}\right) = \frac{1}{4!n^2} $$ It is known that $\sum_{n\ge 1} \frac{1}{n^2} = \frac{\pi^2}{6}.$ Therefore our sum is also convergent.

dnqxt
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