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.

  • $\begingroup$ Do not use series before integration. One integration by parts gives standard integral. $\endgroup$ Commented Apr 23 at 9:57
  • $\begingroup$ 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 → $cos x$. $\endgroup$
    – CiaPan
    Commented Apr 23 at 10:11

1 Answer 1


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

  • $\begingroup$ that was elegant. Thanks. $\endgroup$
    – Debu
    Commented Apr 23 at 10:16

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