# Infinite series question involving integrals

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.

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

• Use the given indication and show where you stuck – HAMIDINE SOUMARE Apr 21 at 21:14
• Agreed. Use the substitution you were given. Do not evaluate the integral. (P.S. In any case, your evaluation is incorrect.) – GEdgar Apr 21 at 21:54
• How? show how that works. – Meghan C Apr 21 at 21:55
• You're right, I've corrected my attempt. – Meghan C Apr 21 at 22:05

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 $$u_{n+1}$$: $$|u_{n+1}| = |\int_{n\pi}^{(n+1)\pi}\frac{-\sin{T}}{T + \pi} dT| \leq \int_{n\pi}^{(n+1)\pi}|\frac{-\sin{T}}{T + \pi}| dT \leq \int_{n\pi}^{(n+1)\pi}\frac{\sin{T}}{T + \pi} dT \lt \int_{n\pi}^{(n+1)\pi}\frac{\sin{T}}{T} dT = u_n \leq |u_n|$$