# Convergence of the sequence $x_{n} = \int_{1}^{n}\frac{\cos t}{t^{2}}$ as n tends to infinity.

The question:

Show that the sequence $$x_{n} = \int_{1}^{n}\frac{\cos t}{t^{2}}$$ as n tends to infinity.

My attempt:

Writing $$\cos t$$ as power series, we get: $$\int_{1}^{n} \frac{1}{t^{2}} - \frac{1}{2!}+\frac{t^{2}}{4!}-\frac{t^{4}}{6!}- \ldots$$ which equals $$\int_{1}^{n} \Big(\frac{1}{t^{2}} - \frac{1}{2!}\Big)$$ + $$\int_{1}^{n}\sum_{m=1}^{\infty}(-1)^{m+1}\frac{t^{2m}}{[2(m+1)]!}$$.

Now Radius of convergence of $$\sum_{m=1}^{\infty} (-1)^{m+1}\frac{y^{m}}{[2(m+1)]!}$$ comes out to be infinity. Thus the integration can be swapped with the summation. Hence we get,

$$\int_{1}^{n} \Big(\frac{1}{t^{2}} + \frac{1}{2!}\Big)$$ + $$\sum_{m=1}^{\infty}\int_{1}^{n}(-1)^{m+1}\frac{t^{2m}}{[2(m+1)]!}$$ which equals

$$1-\frac{1}{n} + \sum_{m=0}^{\infty}(-1)^{m+1}\frac{n^{2m+1}-1}{(2m+1)[(2m+2)]!}$$. Now we observe that

$$\sum_{m=0}^{\infty}\frac{n^{2m+1}-1}{(2m+1)[(2m+2)]!} \leq$$ $$\sum_{m=0}^{\infty}\frac{n^{2m+m}}{(2m+1)[(2m+2)]!}$$ and the radius of convergence of the latter power series again comes out to be infinity. Thus the power series above converges absolutely which implies that

$$1-\frac{1}{n} + \sum_{m=0}^{\infty}(-1)^{m+1}\frac{n^{2m+1}-1}{(2m+1)[(2m+2)]!}$$ converges as n tends to infinity.

I am not very sure if this is correct and this is most elegant way to solve the question. I tried by-parts but it didn't help me. Any help/feedback/insight is appreciated.

• Do not use series before integration. One integration by parts gives standard integral. Commented Apr 23 at 9:57
• You may want to prefix the trigonometric functions' names with a backslash. It will make the LaTeX/MathJax to interpret them as symbols and display in upright font with appropriate spacing: \cos x → $\cos x$ instead of an amorphous blob of italic, variable-like letters: cos x → $cos x$. Commented Apr 23 at 10:11

$$|x_n-x_m| \le \left|\int_n^{m} \frac 1 {t^{2}}dt\right| \to 0$$ so $$(x_n)$$ is Cauchy.