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Can someone help me solve this tricky infinite series problem? I tried to find the indefinite integral of the nth term but my solution didn't make sense at all. enter image description here

I suspect there must be an alternative better approach. Any help would be appreciated.

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  • $\begingroup$ Use the given indication and show where you stuck $\endgroup$ – HAMIDINE SOUMARE Apr 21 at 21:14
  • $\begingroup$ Agreed. Use the substitution you were given. Do not evaluate the integral. (P.S. In any case, your evaluation is incorrect.) $\endgroup$ – GEdgar Apr 21 at 21:54
  • $\begingroup$ How? show how that works. $\endgroup$ – Meghan C Apr 21 at 21:55
  • $\begingroup$ You're right, I've corrected my attempt. $\endgroup$ – Meghan C Apr 21 at 22:05
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First let's try the substitution in evaluating $u_n$: we let $T = t - \pi$, so $du = dt$. The bounds $t = n\pi$ and $t = (n+1)\pi$ get sent to $T = (n-1)\pi$ and $T = n\pi$, respectively. Also note that we get $\sin{t} = \sin(T + \pi) = -\sin{T}$. Then we can write $u_n = \int_{n\pi}^{(n+1)\pi}\frac{\sin{t}}{t}dt = \int_{(n-1)\pi}^{n\pi}\frac{-\sin{T}}{T + \pi} dT$. Thus $u_{n+1} = \int_{n\pi}^{(n+1)\pi}\frac{-\sin{T}}{T + \pi} dT$.

Now note that $|\int f(t)dt| \leq \int|f(t)|dt$, so we can bound $u_{n+1}$: $|u_{n+1}| = |\int_{n\pi}^{(n+1)\pi}\frac{-\sin{T}}{T + \pi} dT| \leq \int_{n\pi}^{(n+1)\pi}|\frac{-\sin{T}}{T + \pi}| dT \leq \int_{n\pi}^{(n+1)\pi}\frac{\sin{T}}{T + \pi} dT \lt \int_{n\pi}^{(n+1)\pi}\frac{\sin{T}}{T} dT = u_n \leq |u_n|$

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  • $\begingroup$ Thanks nilradical1 $\endgroup$ – Meghan C Apr 25 at 14:36

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