I would like to continue the discussion started in this post by offering you a set of even more challenging limits. (As an additional exercise try solving these limits without Riemann Sums) Lets proceed to our first *not so trivial example*.

**Problem 4.** Find the following limit
\[\lim_{n\to\infty}\frac{1}{n}\sqrt[n]{(n+1)(n+2)...(2n)}\]

*Solution.* We can't apply our tactic directly because we have to deal with product rather than sum. However, as many of you can guess, there is a natural way to transform the product into sum.
\[\lim_{n\to\infty}\frac{1}{n}\sqrt[n]{(n+1)(n+2)...(2n)}=\lim_{n\to\infty}\sqrt[n]{\prod^n_{i=1}\left(1+\frac{i}{n}\right)}=\]
\[=\lim_{n\to\infty}{\prod^n_{i=1}\left(1+\frac{i}{n}\right)}^{\frac{1}{n}}=\lim_{n\to\infty}e^{\frac{1}{n}\sum^n_{i=1}\ln\left(1+\frac{i}{n}\right)}\]

Now we see a familiar construction, which can be easily evaluated, namely \[\lim_{n\to\infty}\frac{1}{n}\sum^n_{i=1}\ln\left(1+\frac{i}{n}\right)=\int^1_0\ln(1+x)\,dx=\ln4 - 1.\]

Now returning to the initial problem we get \[\lim_{n\to\infty}\frac{1}{n}\sqrt[n]{(n+1)(n+2)...(2n)}=\lim_{n\to\infty}e^{\frac{1}{n}\sum^n_{i=1}\ln\left(1+\frac{i}{n}\right)}=\] \[=e^{\ln4 - 1}=\boxed{\frac{4}{e}}.\]

Almost the same problem was proposed at National Mathematics Olympiad in Moldova in 2007.

**Problem 5.** *(Moldova, 2007, \(12^{th}\) class)* Calculate the limit
\[\lim_{n\to\infty}\frac{\sqrt[n+1]{(2n+3)(2n+4)...(3n+3)}}{n+1}\]

I also would like to propose another problem from Moldova, this time it's District Mathematics Olympiad, 2013. It's much easier than examples discussed in this post, but still serves a decent technique trainer.

**Problem 6.** *(Moldova, 2013, \(12^{th}\) class)* Find the limit
\[\lim_{n\to\infty}\sum^n_{i=1}\frac{n}{(n+1)\sqrt{n(n+i)}}\]

## Comments

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TopNewestNice post Nicolae! :)

Problem 5:It can be written as:

\(\displaystyle \lim_{n\rightarrow \infty} \exp\left(\frac{1}{n+1}\sum_{i=1}^{n+1}\ln\left(2+\frac{i}{n+1}\right)\right)\)

The above is equivalent to:

\( \displaystyle \exp\left(\int_0^1 \ln(2+x)\,\,dx\right) = \exp\left(\ln\left(\frac{27}{4}\right)-1\right)=\frac{27}{4e}\)

Problem 6:The given expression is equivalent to:

\(\displaystyle \lim_{n\rightarrow \infty} \sum_{i=1}^n \cfrac{1}{n\left(1+\frac{1}{n}\right)\sqrt{1+\frac{i}{n}}} \)

As \(n\rightarrow \infty\), \(1+1/n=1\) and the rest of the expression is same as:

\(\displaystyle \int_0^1 \frac{1}{\sqrt{1+x}}dx=2(\sqrt{2}-1)\) – Pranav Arora · 3 years, 1 month ago

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– Nicolae Sapoval · 3 years, 1 month ago

Good job! Stay tuned for the updates. I plan to write one more post like this and then to end the series on Limits and Integrals with a small testLog in to reply

I eagerly wait for your upcoming posts. – Pranav Arora · 3 years, 1 month ago

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