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I am struggling to find the limit of $(\sqrt{\frac{n}{(n+1)^2(n+2)}}t+\frac{n}{n+1} )^n$ as $n$ goes to $\infty$.

I know that the limit of $(1+\frac{t}{n})^n$ as $n$ goes to $\infty$ equals $e^t$ and that my limit should approach $e^{t-1}$.

However, I am having a hard time using and manipulating the known $(1+\frac{t}{n})^n$ limit to find the value that my limit approaches.

I tried $\lim_{n \to \infty} (\sqrt{\frac{n}{(n+1)^2(n+2)}}t+\frac{n}{n+1} )^n=(\lim_{n \to \infty} (\sqrt{\frac{n}{(n+1)^2(n+2)}}t+\lim_{n \to \infty} \frac{n}{n+1})^n$ to avoid using $\lim_{n \to \infty}(1+\frac{t}{n})^n=e^t$ but I am unsure if it is correct.

If possible, I would prefer a very simple method that just uses properties of limits.

Any help is appreciated, thank you in advanced.

John Doe
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Silvia Rossi
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3 Answers3

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Take natural log. In the derivation below, I use the notation $f(n) \sim g(n)$ to mean $\lim_{n \rightarrow \infty} f(n)/g(n) = 1$, meaning $\lim_{n \rightarrow \infty}f(n) = \lim_{n \rightarrow \infty}g(n)$ if either exists (assume $f,g \neq 0$ for large enough $n$). Since we know $\lim_{x \rightarrow 0} \ln(1 + x)/x = 1$, if $f(n)$ is any function which approaches $0$ as $n \rightarrow \infty$ (but does not equal zero for large enough $n$) we have $\ln(1 + f(n)) \sim f(n)$. Also, we know if $f(n) \sim g(n)$ then $h(n)f(n) \sim h(n)g(n)$ for any function $h$ (again, nonzero for large enough $n$).

\begin{align*}\ln\left[\left(\sqrt{\frac{n}{(n+1)^{2}(n+2)}}t + \frac{n}{n+1}\right)^{n}\right] &= n\ln\left[1 + \frac{1}{n+1}\left(\sqrt{\frac{n}{n+2}}t - 1\right)\right] \\ &\sim n\left[\frac{1}{n+1}\left(\sqrt{\frac{n}{n+2}}t - 1\right)\right] \\ &\sim \sqrt{\frac{n}{n+2}}t - 1 \\ &\sim t - 1 \end{align*}

So the original limit is $e^{t - 1}$

Alex Zorn
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Consider the function $$ f(x)=\frac{1}{x}\log\left(t\sqrt\frac{x^2}{(1+x)^2(1+2x)}+\frac{1}{x+1}\right) $$ (natural logarithm). Then your limit is $$ \lim_{n\to\infty}\exp(f(1/n)) $$ (I just replaced $n$ with $1/x$ and took the logarithm) so we can as well find $\lim_{x\to0^+}f(x)$. For $x>0$ we can pull $x^2$ out of the square root and we just need to find the derivative at $0$ of $$ g(x)=\log\Bigl(\frac{tx}{\sqrt{(1+x)^2(1+2x)}}+\frac{1}{x+1}\Bigr) $$ because $g(0)=0$. This only takes some patience. If $h(x)$ is the argument to the logarithm, then $g(x)=\log(h(x))$ and $g'(x)=h'(x)/h(x)$. Since $h(0)=1$, we reduce to computing $h'(x)$ and $h'(0)$. Now $$ h'(x)=\frac{t\sqrt{(1+x)^2(1+2x)}-tx\dfrac{2(1+x)(1+2x)+2(1+x)^2}{2\sqrt{(1+x)^2(1+2x)}}}{(1+x)^2(1+2x)}-\frac{1}{(x+1)^2} $$ and so $h'(0)=t-1$.

Thus your limit is $e^{t-1}$.

egreg
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In the same spirit as other answers, you could even go beyond the limit.

Starting with $$A=\left(\sqrt{\frac{n}{(n+1)^2(n+2)}}t+\frac{n}{n+1}\right )^n$$

$$\log(A)=n \log\left(\sqrt{\frac{n}{(n+1)^2(n+2)}}t+\frac{n}{n+1}\right )$$ Now, using long division (or better, Taylor series) $$\frac{n}{(n+1)^2(n+2)}=\frac 1 {n^2}-\frac{4}{n^3}+O\left(\frac{1}{n^4}\right)$$ $$\sqrt{\frac{n}{(n+1)^2(n+2)}}=\frac 1 {n}-\frac{2}{n^2}+O\left(\frac{1}{n^3}\right)$$ $$\frac{n}{n+1}=1-\frac 1 {n}+\frac{1}{n^2}+O\left(\frac{1}{n^3}\right)$$ making $$\sqrt{\frac{n}{(n+1)^2(n+2)}}t+\frac{n}{n+1}=1+\frac{t-1}{n}+\frac{1-2 t}{n^2}+O\left(\frac{1}{n^3}\right)$$ $$\log\left(\sqrt{\frac{n}{(n+1)^2(n+2)}}t+\frac{n}{n+1}\right )=\frac{t-1}{n}+\frac{-\frac{t^2}{2}-t+\frac{1}{2}}{n^2}+O\left(\frac{1}{n^3}\right)$$ $$\log(A)=(t-1)+\frac{-\frac{t^2}{2}-t+\frac{1}{2}}{n}+O\left(\frac{1}{n^2}\right)$$ $$A=e^{\log(A)}=e^{t-1}\left(1-\frac{t^2+2t-1 }{2n}\right)+O\left(\frac{1}{n^2}\right)$$ which shows the limit and how it is approached.

Claude Leibovici
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