Small approx

[weary skiff]

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[mellow flare]

my working out:

Please Help!

[frozen path]

approx the trig function > substitute the approx. > simplify the sum > eval the sum > give me the final expression for a_n > do limit eval > simply take the limit lastly

[sly fractal]

cosx isn't smaller than x

[delicate marsh]

I feel as though you need some mildly better approximations

this isn't true at all

Perhaps first we can simplify it a bit

\begin{align*}
\sum_{k=1}^{n-1} \frac{\sin\left( \frac{2k-1}{2n} \pi \right)}{\cos^4\left(\frac{k-1}{n} \frac{\pi}{2} \right)} \leq a_n \leq \sum_{k=1}^{n-1} \frac{\sin\left( \frac{2k-1}{2n} \pi \right)}{\cos^4\left(\frac{k}{n} \frac{\pi}{2} \right)}
\end{align*}

[white raven]

Rion

[delicate marsh]

so you can try with one of them to see what happens

[sly fractal]

do you want other ideas for this problem too?

I don't want to randomly drop unwanted things

[delicate marsh]

Also, something useful to notice:
Let $f_n(x) = \frac{\sin\left(\left(x - \frac{1}{2n} \right) \pi\right)}{\cos^4\left(\frac{\pi}{2}x\right)}$. You can show that $f_n$ is increasing on $(0, 1)$

[white raven]

Rion

[delicate marsh]

so in fact you have a pretty good idea of the largest term of the sum

[mellow flare]

thank you so much sorry i just woke up

[mystic shadow]

@mellow flare has given 1 rep to @delicate marsh

[delicate marsh]

No worries, I hope it helps

[sleek trout]

@mellow flare

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[delicate marsh]

+close