22
$\begingroup$

Problem 9 in the JHMT 2013 Calculus Test asks to evaluate $$\lim_{n\to\infty} \prod_{k=1}^n \frac{2k}{2k-1}\int_{-1}^{\infty} \frac{{\left(\cos{x}\right)}^{2n}}{2^x} \; dx$$ The answer is $\pi\cdot 2^\pi /(2^{\pi}-1)$. How can I show this? I know that the infinite product diverges and the limit cannot be moved into the integral, but I don't know what to do. Maybe I can represent the integral as a summation.

$\endgroup$
  • $\begingroup$ The product is asymptotically $e^{-\sum_k \log (1-\frac{1}{2k}}) = (e^{H_n})^{\frac{1}{2}} \sim \sqrt{n}$ $\endgroup$ – Alex May 23 at 14:20
  • $\begingroup$ @Alex I still don't understand how I can use that to solve the integral $\endgroup$ – Ty. May 23 at 16:35
  • $\begingroup$ @EeyoreHo JHMT 2013 Problem #9 math.jhu.edu/~mathclub/problems/problems2013/Calculus.pdf $\endgroup$ – Ty. May 23 at 20:16
  • 2
    $\begingroup$ Sometimes when I get too full of my self, I look at problems like this with a suggested $5$ minute time to complete it, and I return to normal. $\endgroup$ – Integrand May 23 at 21:12
  • $\begingroup$ It may be that the integral runs from $-\pi$ and not from $-1$. In this case one could simplify a bit the expression by splitting the integral into subintervals, but I was not able to arrive yet to a solution. $\endgroup$ – Thomas May 23 at 21:42
10
$\begingroup$

The identities $$ \frac{2\cdot4\cdot\ldots\cdot (2n)}{1\cdot3\cdot\ldots\cdot(2n-1)}=\frac{\Gamma(n+1)}{\Gamma(n+\tfrac12)}\sqrt{\pi} $$ and Wallis' formula $$ \int^{\frac{\pi}{2}}_0\cos^{2n}x\,dx=\int^{\frac{\pi}{2}}_0\sin^{2n}(x)\,dx=\frac{\Gamma(n+\tfrac12)}{\sqrt{\pi}\Gamma(n+1)}\frac{\pi}{2} $$ will be useful ( a simple derivation of the latter is in Thenard Rinmann's solution). The sequence in your problem can be expressed as $$ I_n:=\frac{\sqrt{\pi}\Gamma(n+1)}{\Gamma(n+\tfrac12)}\int^\infty_{-1}2^{-x}\cos^{2n}x\,dx $$ To make estimates simpler, I only consider the sequence $$ J_n:=\frac{\sqrt{\pi}\Gamma(n+1)}{\Gamma(n+\tfrac12)}\int^\infty_0 2^{-x}\cos^{2n}x\,dx$$ The integral $\int^\infty_0 2^{-x}\cos^{2n}x\,dx$ can be expressed as \begin{aligned} \int^\infty_0 2^{-x}\cos^{2n}x\,dx&=\sum^\infty_{k=0}\int^{(k+1)\pi}_{k\pi}2^{-x}\cos^{2n}x\,dx=\sum^\infty_{k=0}\int^\pi_02^{-(x+ k\pi)}\cos^{2n}(x+k\pi)\,dx \\&=\Big(\sum^\infty_{k=0}2^{-k\pi}\Big)\int^\pi_02^{-x}\cos^{2n}x\,dx=\frac{1}{1-2^{-\pi}}\int^\pi_02^{-x}\cos^{2n}xdx \end{aligned} Here we have used the fact that $\cos(x+k\pi)=(-1)^k\cos(x)$.

Claim I: $\frac{\Gamma(n+1)}{\Gamma(n+\tfrac12)}\sim\sqrt{n}$. This follows from Stirling's approximation: $$\frac{\Gamma(n+1)}{\Gamma(n+\tfrac12)}\sim \frac{n^{n+\tfrac12}e^{-n}}{(n-\tfrac12)^n e^{-(n-\tfrac12)}}$$

Claim II: (Suggested by Raoul below) $\int^{\pi/2}_02^{-x}\cos^{2n}x\,dx=\int^{\pi/2}_0\cos^{2n}x\,dx + o(n^{-1/2})$. To check this, we apply the mean value theorem to get \begin{aligned} \Big|\int^{\pi/2}_0(1-2^{-x})\cos^{2n}x\,dx\Big|\leq \log2\int^{\pi/2}_0x\cos^{2n}x\,dx \end{aligned} The fact that $\frac{\sin x}{x}$ decreases over $[0,\pi]$, implies that $\frac{2}{\pi}x-\sin x\leq0$ on $[0,\pi/2]$ and so, $\frac{x^2}{\pi}+\cos x\leq 1$. Consequently \begin{aligned} \int^{\pi/2}_0x\cos^{2n}x\,dx&\leq \int^{\pi/2}_0x\Big(1-\frac{x^2}{\pi}\Big)^{2n}\,dx\\ &=\frac{\pi}{2}\int^{\pi/4}_0(1-u)^{2n}\,du=\frac{\pi}{2(2n+1)}\Big(1-\big(1-\tfrac{\pi}{4}\big)^{2n+1}\Big) \end{aligned} This proves the claim.

A similar argument shows that \begin{aligned} \int^\pi_{\pi/2}2^{-x}\cos^{2n}x\,dx&=2^{-\pi}\int^0_{-\pi/2}2^{-x}\cos^{2n}(x+\pi)\,dx\\ &=2^{-\pi}\int^{\pi/2}_02^x\cos^{2n}x\,dx=2^{-\pi}\int^{\pi/2}_0\cos^{2n}x\,dx+o(n^{-1/2}) \end{aligned}

It follows that \begin{aligned} J_n&=\frac{1}{1-2^{-\pi}} \frac{\sqrt{\pi}\Gamma(n+1)}{\Gamma(n+\frac12)}\Big((1+2^{-\pi})\int^{\pi/2}_0\cos^{2n}x\,dx+o(n^{-1/2})\Big)\\ &=\frac{2^\pi}{2^\pi-1}(1+2^{-\pi})\frac{\pi}{2}+o(1) \end{aligned}

The contribution of $\frac{\sqrt{\pi}\,\Gamma(n+1)}{\Gamma(n+\tfrac12)}\int^0_{-1}2^{-x}\cos^{2n}x\,dx$ can also be estimated as follows $$ \int^0_{-1}2^{-x}\cos^{2n}x\,dx=\int^1_02^x\cos^{2n}x\,dx=\int^{\tfrac{\pi}{2}}_02^x\cos^{2n}x\,dx-\int^{\frac{\pi}{2}}_12^{x}\cos^{2n}x\,dx$$ The second term is bounded by $$ \int^{\frac{\pi}{2}}_12^x\cos^{2n}x\,dx\leq (\cos 1)^{2n}\Big(\frac{\pi}{2}-1\Big)2^{\pi/2}=o(n^{-1/2}) $$ Consequently \begin{aligned} \frac{\sqrt{\pi}\Gamma(n+1)}{\Gamma(n+\tfrac12)}\int^0_{-1}2^{-x}\cos^{2n}x\,dx&=\left(\frac{\sqrt{\pi}\Gamma(n+1)}{\Gamma(n+\tfrac12)}\int^{\pi/2}_02^{x}\cos^{2n}x\,dx\right) +o(1)\\ &=\left(\frac{\sqrt{\pi}\Gamma(n+1)}{\Gamma(n+\tfrac12)}\Big(\int^{\pi/2}_0\cos^{2n}x\,dx+o(n^{-1/2})\Big)\right) +o(1)\\ &=\frac{\pi}{2}+o(1) \end{aligned}

Putting things together gives $$ I_n=J_n+\frac{\sqrt{\pi}\Gamma(n+1)}{\Gamma(n+\tfrac12)}\int^0_{-1}2^{-x}\cos^{2n}x\,dx=\pi\frac{2^\pi}{2^\pi-1} +o(1) $$

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ You just need a better estimate of the term $$ \int_0^{\pi/2} 2^{-x} \cos^{2n} x \: \mathrm{d} x. $$ For this, check that only the $\cos$ matters, so you just end up with a Wallis integral. More precisely, by mean value theorem, $$ \left | \int_0^{\pi/2} 2^{-x} \cos^{2n} x \: \mathrm{d} x - \int_0^{\pi/2} \cos^{2n} x \: \mathrm{d} x \right | \leq C_1 \int_0^{\pi/2} x \cos^{2n} x \: \mathrm{d} x = o(1/\sqrt{n})$. $$ To see this, cut the integral between $0$ and $n^{-2/3}$ and $n^{-2/3}$ to $\pi/2$. The first part is obvious, and for the second one: $\cos x \leq 1 - C_2 x^2$. $\endgroup$ – Raoul May 24 at 2:02
  • $\begingroup$ @Raoul . By applying the m.v.t. on $f(x)=2^{-x}$ I get $(1-2^{-x})\le ( sup_{z \ in [0,x]}2^{-z}) x=x$. So that $1-2^{-x} \le x$ for $x>0$ which can also be proven by other means. Does this mean that $C_1=1$ ? Oliver Diaz, why do you have $C_1=log(2)$ ? I know the value of the constant is not important but just for clarification. $\endgroup$ – Thomas May 24 at 9:52
  • $\begingroup$ By the way very nice solution putting together a lot of nice ideas. I had just spot the geometric sum than I was blocked :) $\endgroup$ – Thomas May 24 at 10:20
  • $\begingroup$ @Thomas Yes, you are missing a $\ln 2$ in the derivative of $2^{-x}$ to apply the MVT. And much cleaner solution from Oliver to bound the error, nicely done. $\endgroup$ – Raoul May 24 at 14:27
  • $\begingroup$ You are right I forgot the prefactor in deriving $2^{-x}$. Bad mistake from my side but thanks for clarifying. Again, nice solution :) $\endgroup$ – Thomas May 24 at 14:40
6
$\begingroup$

Update

We have \begin{align} &\int_{-1}^\infty \frac{(\cos x)^{2n}}{2^x}\mathrm{d}x\\ =\ & \int_{-1}^0 \frac{(\cos x)^{2n}}{2^x}\mathrm{d}x + \sum_{j=0}^\infty \int_{j\pi}^{(j+1)\pi} \frac{(\cos x)^{2n}}{2^x}\mathrm{d}x\\ =\ & \int_{-1}^0 \frac{(\cos x)^{2n}}{2^x}\mathrm{d}x + \sum_{j=0}^\infty \frac{1}{2^{j\pi}}\int_0^\pi \frac{(\cos x)^{2n}}{2^x}\mathrm{d}x\\ =\ & \int_{-1}^0 \frac{(\cos x)^{2n}}{2^x}\mathrm{d}x + \frac{2^\pi}{2^\pi - 1}\int_0^\pi \frac{(\cos x)^{2n}}{2^x}\mathrm{d}x \\ =\ & \int_{-1}^0 + \frac{2^\pi}{2^\pi - 1} \left(\int_{-1}^{\pi-1} + \int_{\pi-1}^\pi - \int_{-1}^0\right) \\ =\ & \frac{2^\pi}{2^\pi - 1}\int_{-1}^{\pi-1} + \frac{2^\pi}{2^\pi - 1}\int_{\pi-1}^\pi -\frac{1}{2^\pi-1}\int_{-1}^0 \tag{1} \\ =\ & \frac{2^\pi}{2^\pi - 1}\int_{-1}^{\pi-1} + \frac{1}{2^\pi - 1}\int_{-1}^0 -\frac{1}{2^\pi-1}\int_{-1}^0 \tag{2} \\ =\ & \frac{2^\pi}{2^\pi - 1}\int_{-1}^{\pi-1} \frac{(\cos x)^{2n}}{2^x}\mathrm{d}x. \end{align} In (1)(2) we have used
$\int_{\pi-1}^\pi \frac{(\cos x)^{2n}}{2^x}\mathrm{d}x = \frac{1}{2^\pi} \int_{-1}^0 \frac{(\cos y)^{2n}}{2^y}\mathrm{d}y$ (by the substitution $x - \pi = y$).

Also, we have $$\prod_{k=1}^n \frac{2k}{2k-1} = \frac{\Gamma(n+1)\sqrt{\pi}}{\Gamma(n+\frac{1}{2})} \sim \sqrt{n\pi}, \quad \mathrm{as}\quad n\to \infty.$$

Thus, we have \begin{align} &\lim_{n\to \infty} \left(\prod_{k=1}^n \frac{2k}{2k-1}\cdot \int_{-1}^\infty \frac{(\cos x)^{2n}}{2^x}\mathrm{d}x\right)\\ =\ & \lim_{n\to \infty} \left(\sqrt{n\pi}\cdot \frac{2^\pi}{2^\pi - 1}\int_{-1}^{\pi-1} \frac{(\cos x)^{2n}}{2^x}\mathrm{d}x\right)\\ =\ & \pi\frac{2^\pi}{2^\pi-1} \cdot \lim_{n\to \infty} \int_{-1}^{\pi-1} \frac{(\cos x)^{2n}}{2^x} \sqrt{\frac{n}{\pi}}\, \mathrm{d}x. \end{align}

To proceed, we need the following auxiliary results (Facts 1 through 3). The proofs are given at the end.

Fact 1: It holds that, for $-1 \le x \le 1$ and $n \ge 2$, $$\mathrm{e}^{-x^2n} - \frac{1}{n} \le (\cos x)^{2n} \le \mathrm{e}^{-x^2n}.$$

Fact 2: It holds that $$\lim_{n\to \infty} \int_{-1}^1 \frac{\mathrm{e}^{-x^2n}}{2^x} \sqrt{\frac{n}{\pi}}\, \mathrm{d}x = 1.$$

Fact 3: It holds that $$\lim_{n\to \infty} \int_{-1}^1 \frac{\mathrm{e}^{-x^2n} - \frac{1}{n}}{2^x} \sqrt{\frac{n}{\pi}}\, \mathrm{d}x = 1.$$

Now, let us proceed. From Facts 1-3, by the squeeze theorem, we have $$\lim_{n\to \infty} \int_{-1}^1 \frac{(\cos x)^{2n}}{2^x} \sqrt{\frac{n}{\pi}}\, \mathrm{d}x = 1.$$ Also, clearly, we have (since $|\cos x| \le \cos 1 < \frac{3}{5}$ for $1 \le x \le \pi - 1$) $$\lim_{n\to \infty} \int_1^{\pi-1} \frac{(\cos x)^{2n}}{2^x} \sqrt{\frac{n}{\pi}}\, \mathrm{d}x = 0.$$ Thus, we have $$\lim_{n\to \infty} \int_{-1}^{\pi-1} \frac{(\cos x)^{2n}}{2^x} \sqrt{\frac{n}{\pi}}\, \mathrm{d}x = 1.$$

Finally, we have $$\lim_{n\to \infty} \left(\prod_{k=1}^n \frac{2k}{2k-1}\cdot \int_{-1}^\infty \frac{(\cos x)^{2n}}{2^x}\mathrm{d}x\right) = \pi\frac{2^\pi}{2^\pi-1}.$$ We are done.

$\phantom{2}$

Proof of Fact 1: The right inequality is equivalent to $$\ln \cos x \le - \frac{x^2}{2}.$$ The proof is easy and thus omitted.

For the left inequality, clearly, we only need to prove the case when $-\sqrt{\frac{\ln n}{n}} < x < \sqrt{\frac{\ln n}{n}}$. The left inequality is equivalent to $$\ln \left(\mathrm{e}^{-x^2n} - \frac{1}{n}\right) \le 2n\ln \cos x$$ or $$-x^2n + \ln \Big(1 - \frac{\mathrm{e}^{x^2n}}{n}\Big) \le 2n\ln \cos x.$$ Since $\ln (1 - \frac{\mathrm{e}^{x^2n}}{n}) \le - \frac{\mathrm{e}^{x^2n}}{n}$ and $\cos x \ge 1 - \frac{x^2}{2}$, it suffices to prove that $$-x^2n - \frac{\mathrm{e}^{x^2n}}{n} \le 2n\ln \left(1-\frac{x^2}{2}\right).$$ Let $$F(x) = 2n\ln \left(1-\frac{x^2}{2}\right) + x^2n + \frac{\mathrm{e}^{x^2n}}{n}.$$ We have $$F'(x) = \frac{2x}{2-x^2}\left(\mathrm{e}^{x^2n}(2-x^2) - x^2n\right).$$ Since $\mathrm{e}^{x^2n}(2-x^2) - x^2n \ge \mathrm{e}^{x^2n} - x^2n > 0$, we have $F'(x) > 0$ for $0 < x < \sqrt{\frac{\ln n}{n}}$, and $F'(x) < 0$ for $-\sqrt{\frac{\ln n}{n}} < x < 0$. Also, $F(0) > 0$. Thus, $F(x) \ge 0$ for $-\sqrt{\frac{\ln n}{n}} < x < \sqrt{\frac{\ln n}{n}}$. We are done.

$\phantom{2}$

Proof of Fact 2: We have \begin{align} &\lim_{n\to \infty} \int_{-1}^1 \frac{\mathrm{e}^{-x^2n}}{2^x} \sqrt{\frac{n}{\pi}}\, \mathrm{d}x \\ =\ & \lim_{n\to \infty} \int_{-\sqrt{\frac{n}{\pi}}}^{\sqrt{\frac{n}{\pi}}} \mathrm{e}^{-y^2\pi}2^{-y\sqrt{\frac{\pi}{n}}} \mathrm{d}y\\ =\ & \lim_{n\to \infty} \int_{-\sqrt{\frac{n}{\pi}}}^{\sqrt{\frac{n}{\pi}}} \exp\left(-\pi \left(y + \frac{\ln 2}{2\sqrt{\pi n}}\right)^2 + \frac{(\ln 2)^2}{4n}\right) \mathrm{d}y\\ =\ & \lim_{n\to \infty} \exp\left(\frac{(\ln 2)^2}{4n}\right) \int_{-\sqrt{\frac{n}{\pi}} + \frac{\ln 2}{2\sqrt{\pi n}}}^{\sqrt{\frac{n}{\pi}} + \frac{\ln 2}{2\sqrt{\pi n}}} \mathrm{e}^{-\pi z^2} \mathrm{d}z \\ =\ & \int_{-\infty}^\infty \mathrm{e}^{-\pi z^2} \mathrm{d}z\\ =\ & 1. \end{align} We are done.

$\phantom{2}$

Proof of Fact 3: From Fact 2, the desired result follows. We are done.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ (+1) You might consider stating how you arrived at $(1)$. Also, you need to explain how $(2)$ is valid rigoroulsy. You are likely using $\cos(x)=1-\frac12 x^2+O(x^3)$, but not that this is valid for $x$ "near" $0$. $\endgroup$ – Mark Viola May 24 at 17:00
  • $\begingroup$ @MarkViola Thanks. I will update it. $\endgroup$ – River Li May 25 at 0:40
3
$\begingroup$

I think it's simpler to evaluate the integral like this: $$\ $$ We know that by Wallis formula $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(cosx)^{2n}=I_n=\frac{2n-1}{2n}I_{n-1}$$ which on recursive application gives us $$I_n=I_0\prod_{k=1}^n\frac{2k-1}{2k}$$ which gives d$$I_n=\pi\prod_{k=1}^n\frac{2k-1}{2k} \space (\text{as}\space I_0=\pi)$$ and as $n\to\infty$ the value of $$\int_{-1}^{\infty}\frac{(cosx)^{2n}}{2^x}dx$$ will get concentrated near the values where $cosx$ becomes $+1$ or $-1$ and that happens at $0,\pi,2\pi,...$ and the area near other parts of the graph will tend to zero . (I understand that this isn't the most rigorous way to put it, but I believe such ideas are based off the Dominated Convergence Theorem, which I am not very familiar with.) However, answers provided by Oliver Diaz and and River Li give a firm proof for this reasoning. Do look through them for thorough assurance of the idea. For $n=10^{9}$the graph is like this(from desmos)enter image description hereSo, we can write the integral as $$\sum_{k=0}^{\infty}\frac{I_n}{2^{k\pi}}$$ and the total value as $n\to \infty$ becomes equal to $$\prod_{k=1}^n\frac{2k}{2k-1}\int_{-1}^{\infty}\frac{(cosx)^{2n}}{2^x}dx\to \prod_{k=1}^n\frac{2k}{2k-1}\sum_{k=0}^{\infty}\frac{I_n}{2^{k\pi}}=\frac{\pi}{1-2^{-\pi}}=\frac{\pi2^{\pi}}{2^{\pi}-1} $$ and this is valid as long as the lower limit of the integral $$\int_{-1}^{\infty}\frac{(cosx)^{2n}}{2^x}dx$$ more than -$\pi$ and if it's less than $-\pi$ then the lower limit of the summation will become $k=-1$ instead of $k=0$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I didn't think of using Wallis formula, that was good. $\endgroup$ – Ty. May 26 at 14:50
  • 1
    $\begingroup$ You intuitive idea is very good. However more careful analysis is required. Dominated convergence shows that $a_n=\int^0_{-1}2^{-x}\cos^{2n}x\,dx\xrightarrow{n\rightarrow\infty}0$. On the other hand the product $b_n=\prod^n_{k-1}\frac{2k-1}{2n}\sim\sqrt{n}\xrightarrow{n\rightarrow\infty}\infty$. That is why a careful analysis of the rate of convergence of the factor $a_n$ is needed to balanced the competing factor $b_n$ that is growing. $\endgroup$ – Oliver Diaz May 26 at 22:21
1
$\begingroup$

Firstly split it up into two parts: $$\prod_{k=1}^n\frac{2k}{2k-1}=\frac{2.4.6.8...2n}{1.3.5.7.(2n-1)}=\frac{2^nn!\times2^{n-1}(n-1)!}{(2n-1)!}=\frac{2^{2n-1}n!(n-1)!}{(2n-1)!}=\frac{2^{2n-1}(n!)^2}{n(2n-1)!}$$ now the integral: $$I_n=\int_{-1}^\infty\frac{(\cos x)^{2n}}{2^x}dx$$ $$I_n(a)=\int_{-1}^\infty e^{-ax}\cos^{2n}xdx$$ and we know that: $$\cos^{2n}x=\frac{(e^{ix}+e^{-x})^{2n}}{2^{2n}}$$ and: $$(e^{ix}+e^{-ix})^{2n}=\sum_{r=0}^{2n}{{2n}\choose{r}}e^{(2n-r)ix}e^{-rix}=\sum_{r=0}^{2n}{{2n}\choose{r}}e^{(2n-2r)ix}$$ so our integral becomes: $$I_n(a)=\int_{-1}^\infty\sum_{r=0}^{2n}{{2n}\choose{r}}e^{(2n-2r)ix-ax}dx=I_n(a)=\int_{-1}^\infty\sum_{r=0}^{2n}{{2n}\choose{r}}e^{(2i(n-r)-a)x}dx$$ assuming we can interchange the integral and summation and allowing $-b=2i(n-r)-a$ we get: $$I_n(a)=\sum_{r=0}^{2n}{{2n}\choose{r}}\int_{-1}^\infty e^{-bx}dx=\sum_{r=0}^{2n}{{2n}\choose{r}}\left[\frac{-e^{-bx}}{b}\right]_{-1}^\infty=\sum_{r=0}^{2n}{{2n}\choose{r}}\frac{e^b}{b}$$ $$I_n(a)=\sum_{r=0}^{2n}{{2n}\choose{r}}\frac{e^{a-2i(n-r)}}{a-2i(n-r)}$$ If we bring it all together we get: $$L=\lim_{n\to\infty}\frac{2^{2n-1}(n!)^2}{n(2n-1)!}\sum_{r=0}^{2n}{{2n}\choose{r}}\frac{e^{\ln(2)-2i(n-r)}}{\ln(2)-2i(n-r)}$$ and we know that: $${2n\choose r}=\frac{(2n)!}{r!(2n-r)!}=\frac{2^nn!}{r!(2n-r)!}$$ so: $$L=\lim_{n\to\infty}\frac{2^{3n}(n!)^3}{n(2n-1)!}\sum_{r=0}^{2n}\frac{e^{-2i(n-r)}}{\ln(2)-2i(n-r)}\times\frac{1}{r!(2n-r)!}$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I guess that atrocious limit converges to $\pi\cdot\frac{2^{\pi}}{2^{\pi}-1}$, but I feel like there is some easy approach that we're overlooking. This is a part of a calculus competition for high schoolers, and the answer was only accepted as $\pi\cdot\frac{2^{\pi}}{2^{\pi}-1}$. $\endgroup$ – Ty. May 23 at 18:06
  • 1
    $\begingroup$ With respect, I don't feel this is a satisfactory answer. It is not at all obvious the final limit is what the OP claims it is. $\endgroup$ – Integrand May 23 at 21:04
  • $\begingroup$ @OliverDiaz I rewrote the summation part in the question because I didn't think it was what OP was really trying to ask, but now I see I was a bit hasty on that. $\endgroup$ – Bladewood May 24 at 15:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.