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  1. Find the limit $$\lim_{n \rightarrow \infty} \int _0^n (1 - \frac{x}{n})^n \log (2 + \cos(\frac{x}{n})) dx$$ and justify the answer.

I think that the Dominated Convergence Theorem can be applied to this problem.

Since $$\int _0^n (1 - \frac{x}{n})^n \log (2 + \cos(\frac{x}{n})) dx = \int _0^\infty (1 - \frac{x}{n})^n \log (2 + \cos(\frac{x}{n}))1_{[0, n]}(x) dx,$$ if I can find the dominating function, I will then apply DCT.

After applying DCT, we will have $$\lim \int _0^\infty (1 - \frac{x}{n})^n \log (2 + \cos(\frac{x}{n}))1_{[0, n]}(x) dx = \int _0^\infty \lim (1 - \frac{x}{n})^n \log (2 + \cos(\frac{x}{n}))1_{[0, n]}(x) dx = \int _0^\infty e^{-x} \log (3) dx = \log(3)e^{-x}|_{x=\infty, x=0} = - \log (3).$$

But I am not sure if $$\Big| (1 - \frac{x}{n})^n \log (2 + \cos(\frac{x}{n})) \Big| \leq 3e^{-x}$$ with $3e^{-x}$ integrable on $[0, \infty).$ Can someone check if my dominating function is alright ?

  1. Prove that $$\sum_{k = 1}^\infty \frac{1}{(p + k)^2} = - \int_0^1 \frac{x^p}{1-x}\log (x) dx$$ for $p>0.$ (This problem is allow to use the Fundamental Theorem of Calculus)

I suppose that I should apply FTC with DCT. But I have no ideas about doing this, any hints please ?

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    $\begingroup$ I think your dominating function for $1$ holds. $\endgroup$ – Moya Oct 6 '15 at 3:43
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As Moya stated in the comment section, your dominating function is correct. Indeed it can be shown (and it is in fact well known) that $$\Big(1 - \frac xn\Big)^{n} \uparrow e^{-x}.$$

For the other term notice that the argument in the $\log$ is allowed to range from in $[1,3]$ and $\log x$ in this interval is a positive increasing function, so that you can estimate it with $\log 3$ (you have $3$ multiplying the exponential which is fine, the estimate doesn't have to be sharp.)

Notice that you missed a minus sign when evaluating the integral, the result indeed should be $\log 3$

To answer your second question, notice that for $x \in (0,1)$ we can write $\frac{1}{1 - x}$ as a geometric series, indeed

\begin{align} - \int_0^1\frac{x^p}{1 - x}\log x\,dx = &\ - \int_0^1\sum_{k = 0}^{\infty}x^{k + p}\log x\,dx \\ = &\ \sum_{k = 0}^{\infty} \int_0^1-x^{k + p}\log x\,dx \tag 1\\ = &\ \sum_{k = 0}^{\infty}\frac{1}{(k + p + 1)^2} \\ = &\ \sum_{k = 1}^{\infty}\frac{1}{(k + p)^2} \end{align}

Notice that the crucial step here is $(1)$: to move the sum out of the integral you need to apply the Monotone convergence theorem to the partial sums. This can be done since $-x^{k + p}\log x$ is positive making the partial sums a nonnegative increasing sequence.

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    $\begingroup$ Nice solution ...+1 $\endgroup$ – Mark Viola Oct 6 '15 at 14:00
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Since we know that $\left(1-\frac xn\right)^n$ is an increasing function of $n$ with limit $e^{-x}$ for $x>0$, then clearly $\left(1-\frac xn\right)^n \le e^{-x}$ for $x\ge 0$.

And since $-1\le \cos x\le 1$, then clearly $\log \left(2+\cos \left(\frac xn\right)\right)\le \log 3$

Putting these together gives

$$\left(1-\frac xn\right)^n\log \left(2+\cos \left(\frac xn\right)\right)\le \log (3)e^{-x}$$

and we have the dominating function since the indicator function is bounded by $1$.

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  • $\begingroup$ there you go, +1 $\endgroup$ – Giovanni Oct 6 '15 at 5:21
  • $\begingroup$ @Dr.MV Thank you very much for your help. I check the one up vote for your comment. I mostly ask the question than answer the question, so I do not quite know the importance of having up vote and fame point. Please keep on helping solving math problems :). $\endgroup$ – Both Htob Oct 6 '15 at 12:07
  • $\begingroup$ You're welcome. It was my pleasure $\endgroup$ – Mark Viola Oct 6 '15 at 14:00

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