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Calculus Level 3

$\large \lim_{n \to \infty} \frac{{e} \times e^{\frac{1}{3}} \times e^{\frac{1}{5}} \times e^{\frac{1}{7}} \times \ldots \times e^{\frac {1}{2n+1}}} {e^{\frac{1}{2}} \times e^{\frac{1}{4}} \times e^{\frac{1}{6}} \times \ldots \times e^{\frac {1}{2n}} } = \ ?$

Try this one

The answer is 2.

2 solutions

Tanishq Varshney
Apr 4, 2015

$\large{e^{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}...}}$

$ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-....$

put $x=1$

$e^{ln2}$

$\boxed{2}$

Simple but very nice solution :D

Paul Ryan Longhas - 6 years, 2 months ago

oh yea I forgot about the expansion of Ln(1+x). I tried multiplying by the product on the denominator to give: $\lim_{n \rightarrow\infty} \frac{\displaystyle\prod_{k=1}^{2n+1} e^{\frac{1}{k}} }{\displaystyle\prod_{k=1}^{n} e^{\frac{1}{k}} }= \lim_{n \rightarrow\infty } \displaystyle\prod_{k = n+1}^{2n} e^{\frac{1}{k}}$ $= e^u \ where \ u = \lim_{n \rightarrow\infty} \displaystyle\sum_{k=1}^{n} \frac{1}{n+k} = \frac{1}{n+1} +\frac{1}{n+2} +...+ \frac{1}{2n}$ But I couldn't figure out the last step. Do you know how I could show that $\lim_{n \rightarrow\infty} \displaystyle\sum_{k=1}^{n} \frac{1}{n+k} = Ln 2 \ \ ?$

Curtis Clement - 6 years, 2 months ago

That's equal to $H_{2n}-H_n$ , where $H_n=1+\frac12+\dotsb+\frac1n$ . Since $H_n\approx\ln n+\gamma$ , we have $\lim_{n\to\infty}(H_{2n}-H_n)$ $=\lim_{n\to\infty}((\ln 2n+\gamma)-(\ln n+\gamma))$ $=\lim_{n\to\infty}(\ln2n-\ln n)$ $=\ln2$ .

Or, alternatively, you can show that $\sum_{k=1}^n\frac1{n+k}=1-\frac12+\dotsb-\frac1{2n}$ , and use the series for $\ln2$ . (Hint: Show that they're both equal to $H_{2n}-H_n$ .)

Or, even more alternatively, use the ideas used in my answer, using the squeeze theorem.

Akiva Weinberger - 6 years, 1 month ago

Akiva Weinberger
May 9, 2015

Tanishq Varshney has given the "obvious" solution. I'll add another one. This one uses no infinite series, no integrals, no derivatives. In fact, the only "calculus-y" thing is the squeeze theorem at the very end!

Multiplying top and bottom by $e^{\frac12}\times\dotsb\times e^{\frac1{2n}}$ and rearranging, we find that this is equal to: $\lim_{n\to\infty}e^{\frac1{n+1}+\frac1{n+2}+\frac1{n+3}+\dotsb+\frac1{2n+1}}$ Now, recall that $e^x\ge x+1$ for all $x$ . ( Helpful graph )

Letting $x=\frac1{n+1},\frac1{n+2},$ etc., and multiplying them together, we get the following: $e^{\frac1{n+1}+\dotsb+\frac1{2n+1}}\ge\frac{n+2}{n+1}\times\frac{n+3}{n+2}\times\dotsb\times\frac{2n+2}{2n+1}=\frac{2n+2}{n+1}=2$ Thus, we get a lower bound of $2$ . In addition, take $e^x\ge x+1$ again, and let $x=-\frac1{n+1},-\frac1{n+2}$ , etc. This gives:

$e^{-\frac1{n+1}-\dotsb-\frac1{2n+1}}\ge\frac n{n+1}\times\frac{n+1}{n+2}\times\dotsb\times\frac{2n}{2n+1}=\frac n{2n+1}$

Taking the reciprocal: $e^{\frac1{n+1}+\dotsb+\frac1{2n+1}}\le\frac{2n+1}n=2+\frac1n$

So, we now have: $2\le e^{\frac1{n+1}+\dotsb+\frac1{2n+1}}\le2+\frac1n$ The squeeze theorem tells us that the limit is $2$ .

Awfully Familiar

The answer is 2.

2 solutions

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