Convergence of sequence

[MaxL0110]

Hello, i need find partial sum and verify convergence of this sequence. I have idea to apply euler formula. Early i solved similiar problem but this square, i dont know. \begin{equation}
\sum_{n=1}^{\infty} \cos ^{2}(n x)
\end{equation}
Help please

 

[Deleted member 4993]

Hello, i need find partial sum and verify convergence of this sequence. I have idea to apply euler formula. Early i solved similiar problem but this square, i dont know. \begin{equation}
\sum_{n=1}^{\infty} \cos ^{2}(n x)
\end{equation}
Help please

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem

 

[MaxL0110]

Please show us what you have tried and exactly where you are stuck.

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem

\begin{equation}
\begin{array}{l}
\left(\frac{e^{i x k}+e^{-i x k}}{2}\right)^{2}=\cos ^{2} k x \\
\text { I tried to apply same approach, like } \sum_{n=1}^{\infty} \sin (n x)=\operatorname{Im} \sum_{n=1}^{\infty} e^{i n x} \\
\text { Then used sum of geometrical progression t } \frac{1-t^{n}}{1-t} \text { . } \\
\text { But I think this method will not work here. } \\
\text { I’m stuck here. }
\end{array}
\end{equation}

 

[skeeter]

this power reduction identity may help …

[MATH]\cos^2(nx) = \frac{1+\cos(2 \cdot nx)}{2}[/MATH]

 

[MaxL0110]

[MATH][/MATH]

this power reduction identity may help …
Thank you, i tried it yesterday. I cant understand if i make split into two sums. One sum of cos function [MATH]\cos(2\cdot nx) [/MATH], second of infinite sum of [MATH]\frac{1}{2} [/MATH][MATH]\cos^2(nx) = \frac{1+\cos(2 \cdot nx)}{2}[/MATH]How to express second sum in partial sum. Sry for my english

 

[lex]

[MATH]\begin{equation} \begin{array}{l} \text { I tried to apply same approach, like }\\ \sum_{n=1}^{\infty} \sin (n x)=\operatorname{Im} \sum_{n=1}^{\infty} e^{i n x} \\ \text { Then used sum of geometrical progression t } \frac{1-t^{n}}{1-t} \text { . } \\ \text { But I think this method will not work here. } \\ \text { I’m stuck here. } \end{array} \end{equation}[/MATH]

That's grand, you can do the same, using @skeeter 's hint.
We want (the partial sum) [MATH]\dfrac{1}{2}\sum_{k=1}^{n} (1+\cos(2kx))[/MATH]which is [MATH]\hspace2ex \dfrac{n}{2}+ \dfrac{1}{2} Re\left(\sum_{k=1}^{n} e^{2ikx}\right)\hspace2ex[/MATH] (*)

[MATH]\begin{align*} \sum_{k=1}^{n} e^{2ikx} & = e^{2ix} \dfrac{e^{2inx}-1}{e^{2ix}-1}\\ & =\dfrac{e^{2inx}-1}{1-e^{-2ix}} \end{align*}[/MATH]
Write them out in terms of cos and sin, make the bottom real, find the real part and then use this in (*) to find the partial sum.

You can then make a judgement about convergence.

 

[Steven G]

Recall that a necessary condition for series to converge is that the nth term goes to 0. Go with that and the hints above.

 

[Steven G]

Can you think of a sub-series (is that even a term?) where the terms are not going to 0, ie say always greater than say 1/2?
Can't 2pi be approximated by a decimal number??

 

[Deleted member 4993]

Recall that a necessary condition for series to converge is that the nth term goes to 0. Go with that and the hints above.

I had thought that that statement refers to "infinite series" - not "partial series".

 

[Steven G]

I had thought that that statement refers to "infinite series" - not "partial series".

What did I say wrong??