Problem 15 *

fwiw there are two problems here, 15 and 16.

1 Like

Ah thanks, will tidy up

we have

\begin{align} \lim_{n\to\infty}\int_1^n\frac{n}{1+x^n}dx &= \lim_{n\to\infty}\left(\int_1^n n(1+x^n)^{-1}dx\right) \\ &= \lim_{n\to\infty}\left(\int_1^n nx^{-n}(1+x^{-n})^{-1}dx\right) \end{align}

now from the limits of integration we know, taking n>1, that x\in[1,n]\implies x^{-n}\in[n^{-n},1] so that |x^{-n}|<1 \ \ \forall \ n.

\begin{align} \implies\lim_{n\to\infty}\left(\int_1^n nx^{-n}(1+x^{-n})^{-1}dx\right) &=\lim_{n\to\infty}\left(\int_1^n nx^{-n}\left(\sum_{k=0}^{\infty}\frac{\prod_{m=0}^{k-1}(-1-m)}{k!}(x^{-n})^k\right)dx\right) \\ &= \lim_{n\to\infty}\left(\int_1^n nx^{-n}\left(\sum_{k=0}^{\infty}(-1)^{(k-1)-(0)+1}\frac{\prod_{m=0}^{k-1}(1+m)}{k!}x^{-kn}\right)dx\right) \\ &= \lim_{n\to\infty}\left(\int_1^n nx^{-n}\left(\sum_{k=0}^{\infty}(-1)^{k}\frac{\prod_{m=1}^{k}m}{k!}x^{-kn}\right)dx\right) \end{align}

by the binomial theorem. we know \displaystyle \prod_{m=1}^{k}m=k! so we can simplify this further.
we then obtain

\begin{align} \lim_{n\to\infty}\left(\int_1^n nx^{-n}\left(\sum_{k=0}^{\infty}(-1)^{k}\frac{\prod_{m=1}^{k}m}{k!}x^{-kn}\right)dx\right) &=\lim_{n\to\infty}\left(\int_1^n nx^{-n}\left(\sum_{k=0}^{\infty}(-1)^{k}\frac{k!}{k!}x^{-kn}\right)dx\right) \\ &=\lim_{n\to\infty}\left(n\int_1^n \sum_{k=0}^{\infty}(-1)^{k}x^{-(k+1)n}dx\right) \end{align}

now we can interchange the sum and integral due to convergence of the sum.

\begin{align} \implies\lim_{n\to\infty}\left(n\int_1^n \sum_{k=0}^{\infty}(-1)^{k}x^{-(k+1)n}dx\right) &=\lim_{n\to\infty}\left(n\sum_{k=0}^{\infty} (-1)^{k}\int_1^nx^{-(k+1)n}dx\right) \\ &=\lim_{n\to\infty}\left(n\sum_{k=0}^{\infty} (-1)^{k}\left[\frac{x^{1-(k+1)n}}{1-(k+1)n}\right]_1^n\right) \\ &=\lim_{n\to\infty}\left(n\sum_{k=0}^{\infty} (-1)^{k}\left(\frac{n^{1-(k+1)n}-1}{1-(k+1)n}\right)\right) \\ &=\lim_{n\to\infty}\left(\sum_{k=0}^{\infty} (-1)^{k-1}\left(\frac{n^{2-(k+1)n}-n}{(k+1)n-1}\right)\right) \\ &=\sum_{k=0}^{\infty} (-1)^{k-1}\left(\lim_{n\to\infty}\left(\frac{n^{2-(k+1)n}-n}{(k+1)n-1}\right)\right) \\ &=\sum_{k=0}^{\infty} (-1)^{k-1}\left(\lim_{n\to\infty}\left(\frac{n^{1-(k+1)n}-1}{k+1-\frac 1 n}\right)\right) \\ &=\sum_{k=0}^{\infty} (-1)^{k-1}\left(\lim_{n\to\infty}\left(\frac{n^{1-(k+1)n}}{k+1-\frac 1 n}\right)-\lim_{n\to\infty}\left(\frac{1}{k+1-\frac 1 n}\right)\right) \\ &=\sum_{k=0}^{\infty} (-1)^{k-1}\left(\lim_{n\to\infty}\left(\frac{n^{1-(k+1)n}}{k+1-\frac 1 n}\right)-\frac{1}{k+1}\right) \end{align}

now we must deal with the remaining limit:

\begin{align} \lim_{n\to\infty}\left(\frac{n^{1-(k+1)n}}{k+1-\frac 1 n}\right) &=\lim_{n\to\infty}\left(\frac{\frac{n}{n^{(k+1)n}}}{\frac{(k+1)n-1}{n}}\right) \\ &=\lim_{n\to\infty}\left(\frac{n^2}{n^{(k+1)n}((k+1)n-1)}\right) \\ &=\lim_{n\to\infty}\left(\frac{n^2}{(k+1)n^{(k+1)n+1}}\right) \\ &=\lim_{n\to\infty}\left(\frac{1}{(k+1)n^{(k+1)n-1}}\right) \\ &=\lim_{n\to\infty}\left(\frac{1}{(k+1)n^{(k+1)n}}\right) \end{align}

because (k+1)nÂ±1\to(k+1)n as n\to\infty. so now by the algebra of limits we have

\begin{align} \lim_{n\to\infty}\left(\frac{1}{(k+1)n^{(k+1)n}}\right) &= \frac{1}{(k+1)\lim_{n\to\infty}(n^{(k+1)n})} \\ &= \frac{1}{(k+1)(\lim_{n\to\infty}(n^{n}))^{k+1}} \end{align}

but n^n\to\infty as n\to\infty so as n\to\infty, we have

\begin{align} (k+1)\left(\lim_{n\to\infty}(n^{n})\right)^{k+1}&\to\infty \\ \implies \frac{1}{(k+1)(\lim_{n\to\infty}(n^{n}))^{k+1}}&\to0 \\ \implies \lim_{n\to\infty}\left(\frac{n^{1-(k+1)n}}{k+1-\frac 1 n}\right)&=0 \end{align}
\begin{align} \therefore \sum_{k=0}^{\infty} (-1)^{k-1}\left(\lim_{n\to\infty}\left(\frac{n^{1-(k+1)n}}{k+1-\frac 1 n}\right)-\frac{1}{k+1}\right) &= \sum_{k=0}^{\infty} (-1)^{k-1} \left(0-\frac{1}{k+1}\right) \\ &= \sum_{k=0}^{\infty}\frac{(-1)^k}{k+1} \\ &= \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}(1)^k \\ &=\ln(1+1) \\ &=\ln2 \end{align}

due to the maclaurin series expansion of \displaystyle\ln(1+x)=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}x^k
if you replaced the upper limit with \pi, or in fact generally any a\in\mathbb{R}^+ where a>1, then i claim that you would still get \ln2 (very nice). you would have, very similarly

\begin{align} \lim_{n\to\infty}\int_1^a\frac{n}{1+x^n}dx &=\lim_{n\to\infty}\left(n\sum_{k=0}^{\infty} (-1)^{k}\left[\frac{x^{1-(k+1)n}}{1-(k+1)n}\right]_1^a\right) \\ &=\lim_{n\to\infty}\left(n\sum_{k=0}^{\infty} (-1)^{k}\left(\frac{a^{1-(k+1)n}-1}{1-(k+1)n}\right)\right) \\ &=\lim_{n\to\infty}\left(\sum_{k=0}^{\infty} (-1)^{k-1}\left(\frac{na^{1-(k+1)n}-n}{(k+1)n-1}\right)\right) \\ &=\sum_{k=0}^{\infty} (-1)^{k-1}\left(\lim_{n\to\infty}\left(\frac{na^{1-(k+1)n}-n}{(k+1)n-1}\right)\right) \\ &=\sum_{k=0}^{\infty} (-1)^{k-1}\left(\lim_{n\to\infty}\left(\frac{a^{1-(k+1)n}}{k+1-\frac 1 n}\right)-\frac{1}{k+1}\right) \end{align}

again considering the remaining limit,

\begin{align} \lim_{n\to\infty}\left(\frac{a^{1-(k+1)n}}{k+1-\frac 1 n}\right)&=\lim_{n\to\infty}\left(\frac{\frac{a}{a^{(k+1)n}}}{-\frac 1 n}\right) \\ &= -a\lim_{n\to\infty}\left(\frac{\frac{1}{a^{(k+1)n}}}{\frac 1 n}\right) \\ &= -a\lim_{n\to\infty}\left(\frac{n}{a^{(k+1)n}}\right) \end{align}

we now have a limit in an indeterminate form of \displaystyle \frac \infty \infty which we can evaluate using lâ€™hĂ´pitalâ€™s rule

\begin{align} -a\lim_{n\to\infty}\left(\frac{n}{a^{(k+1)n}}\right)&=-a\lim_{n\to\infty}\left(\frac{\frac{d}{dn}(n)}{\frac{d}{dn}(a^{(k+1)n})}\right) \\ &=-a\lim_{n\to\infty}\left(\frac{1}{(k+1)a^{(k+1)n}\ln a}\right) \\ &=-\frac{a}{(k+1)\ln a}\lim_{n\to\infty}\left(a^{-(k+1)n}\right) \\ &= 0 \end{align}

because clearly, a^{-(k+1)n}\to0 as n\to\infty

\begin{align} \therefore \sum_{k=0}^{\infty} (-1)^{k-1}\left(\lim_{n\to\infty}\left(\frac{a^{1-(k+1)n}}{k+1-\frac 1 n}\right)-\frac{1}{k+1}\right) &=\sum_{k=0}^{\infty} (-1)^{k-1} \left(0-\frac{1}{k+1}\right) \\ &= \ln 2 \end{align}

as before.

f
xD

1 Like

anyone have a more simple solution? think i might have overcomplicated it

I will have a look and let you know if I get anything. Your answer to the second part looks nice enough but Iâ€™ll see if thereâ€™s any nice simplification to the work done in the first part.

First look: you overcomplicated a bit using the binomial theorem. You could have just noted that:

\displaystyle \frac 1 {1 + \left(\frac 1 x\right)^n} = \sum_{k = 0}^\infty \frac {(-1)^k} {x^{kn}}

because:

\displaystyle \frac 1 {1 - x} = \sum_{k = 0}^\infty x^n

for |x| < 1.

Otherwise my idea basically follows yours, I just wouldâ€™ve used fewer lines.

ah true, thanks for noting that. forgot to add that on the second time (i wrote it up once more above my latest post, which is deleted now), the first time i had it as a â€śsimpler alternativeâ€ť lol
i tried and put in additional steps iâ€™d mentally skip for ease of understanding but it seems i kind of went overboard