Limit $ \lim_{n \rightarrow \infty}\sqrt{(1+\frac{1}{n}+\frac{1}{n^3})^{5n+1}} $

Question

What's the limit of this? $$ \lim_{n \rightarrow \infty}\sqrt{\left(1+\frac{1}{n}+\frac{1}{n^3}\right)^{5n+1}} $$

Can't find a way to solve it, I tried to use something looks like $(1+1/n)^n$ but didn't succeed.

Many thanks!

Answer 1

Let $$ a_n = \left(1 + \frac 1n + \frac 1{n^3}\right)^n. $$ To use that limit, we squeeze it like so: $$ \left(1 + \frac 1n\right)^n \leqslant a_n \leqslant \left(1 + \frac {n^2+1}{n^3}\right)^{n^3/(n^2+1)}, $$ so $\lim a_n = \mathrm e$. Then the original limit is $\mathrm e^{5/2}$ by breaking the expression into $$ \sqrt {a_n ^5 \cdot \left(1+\frac 1n + \frac 1{n^3}\right)}, $$ and the expression in the () has the limit $1$.

UPDATE

The proof is actually not complete. If only the limit $\lim (1+\frac 1n)^n$ is allowed to use, then we need to justify $$ \lim_n \left(1+\frac {n^2+1}{n^3}\right)^{n^3/(n^2+1)} = \mathrm e. $$ Then we still need to squeeze this. Let this sequence be $b_n$, then $$ b_n = \left(1 + \frac 1{\frac {n^3}{n^2+1}}\right)^{n^3/(n^2+1)}, $$ note that $$ \frac {n^2+1}{n^3} \leqslant \frac {n^2+n}{n^3} = \frac {n+1}{n^2} \leqslant \frac {n+1}{n^2-1} = \frac 1{n-1}, $$ and also $$ \frac {n^3}{n^2+1} \leqslant \frac {n^3+n}{n^2+1} = n, $$ so $$ \left(1+\frac 1n \right)^n \leqslant a_n \leqslant b_n \leqslant \left(1+\frac 1{n-1}\right)^n = \left(1+\frac 1{n-1}\right)^{n-1}\cdot \left(1 +\frac 1{n-1}\right), $$ then take the limit $n\to \infty$, we have $$ \lim a_n = \lim b_n =\mathrm e. $$

Remark

This seems complicated, so actually you could show that $$ \lim_n \left(1 + \frac 1 {f(n)}\right)^{f(n)} =\mathrm e $$ where $\lim f(n) = +\infty$ after you learning the limit of functions, i.e. the well known $$ \lim_{x\to +\infty} \left(1+\frac 1x\right)^x = \mathrm e $$ to make things a bit easier by invoking the relationship between limits of sequences and functions. This answer is a demo when there is some restrictions on the tools you are allowed to use.

Answer 2

You may use the fact:

• $(1+x_n)^{\large \frac{1}{x_n}} \stackrel{n \to \infty}{\longrightarrow} e$ for any sequence $(x_n)$ with $0<x_n \stackrel{n \to \infty}{\longrightarrow} 0$

It follows

$$\begin{eqnarray*} \sqrt{\left(1+\frac{1}{n}+\frac{1}{n^3}\right)^{5n+1}} & = & \sqrt{\left(1 + \frac{n^2+1}{n^3}\right)^{\large\frac{n^3}{n^2+1}\cdot \underbrace{\large \frac{(5n+1)(n^2+1)}{n^3}}_{\large \stackrel{n \to \infty}{\longrightarrow} 5}}} \\ & \stackrel{n \to \infty}{\longrightarrow} & \sqrt{e^5} \end{eqnarray*}$$

Answer 3

Short answer: (correct but not 100% rigorous, see @xbh)

The term $1/n^3$ is negligible compared to $1/n$, and $1$ is negligible compared to $5n$. Then your limit is the same as that of

$$\left(\left(1+\frac1n\right)^{1/n}\right)^{5/2}.$$

---

With a little more trickery, you can write

$$\frac1m=\frac1n+\frac1{n^3}$$ and it is not a big deal to show

$$n=m+o(m).$$

Now you have

$$\sqrt{\left(1+\frac1m\right)^{5m+o(m)+1}}$$ where only the first exponent brings a non-unit contribution.

Answer 4

Hint:

$$\lim_{n\to\infty}\left(1+\dfrac{n^2+1}{n^3}\right)^{(5n+1)/2}$$

$$=\left(\lim_{n\to\infty}\left(1+\dfrac{n^2+1}{n^3}\right)^{n^3/(n^2+1)}\right)^{\lim_{n\to\infty}\dfrac{(5n+1)(n^2+1)}{2n^3}}$$

As $\lim_{n\to\infty}\dfrac{n^2+1}{n^3}=0,$ the inner limit converges to $e$

$\lim_{n\to\infty}\dfrac{(5n+1)(n^2+1)}{2n^3}=\lim_{n\to\infty}\dfrac{\left(5+\dfrac1n\right)\left(1+\dfrac1{n^2}\right)}2=?$

Answer 5

Rewrite: $$\sqrt{\left(1+\frac{1}{n}+\frac{1}{n^3}\right)^{5n+1}}=\exp\left(\tfrac{5n+1}{2}\log\left(1+\tfrac{1}{n}+\tfrac{1}{n^3}\right)\right)$$ and $\log(1+x) \approx x$ for small $x$, so: $$\lim_{n \to \infty}\sqrt{\left(1+\frac{1}{n}+\frac{1}{n^3}\right)^{5n+1}}=\lim_{n \to \infty}\exp\left(\tfrac{5n+1}{2}\left(\tfrac{1}{n}+\tfrac{1}{n^3}\right)\right)=e^{5/2}$$

Answer 6

It's $$e^{5/2} .$$ I've been involved in a few posts in which the query is like that of this one, recently, the subject being one of my favourite ones.

It scarcely matters about the square-root, really: that doesn't affect the limiting process itself here atall. A crucial point though is that in a limit such as this, the terms of higher order than linear in $1/n$ become negligible. This can be shown in various ways, such as by analysing the terms in a generalised expansion combinatorially. I have discussed this elsewhere; and I think a repetition of any more than a little of it would be a tad redundant here. But I'll say that it's a bit like the way the terms beyond the linear ones fade out in the derivation of the derivative of a function - but that here it's 'cast' in a multiplicative form.

Also the constant (with respect to $n$) in the exponent becomes negligible: this can be seen by noting that leaving it in would be the same as leaving a finite power of the bracket multiplying the whole thing; And the limit of a finite power of it as $n\to\infty$ is certainly 1.

So we are left with $$\lim_{n\to\infty}\left(1+{1\over n}\right)^{5n\over2}$$ This is now just the standard limit form of the exponential function $$\lim_{n\to\infty}\left(1+{x\over n}\right)^n$$$$=$$$$\lim_{n\to\infty}\left(1+{1\over n}\right)^{nx}$$$$=$$$$\lim_{n\to\infty}\left(1+{\sqrt{x}\over n}\right)^{n\sqrt{x}}, $$ or whatever.

The reason I have written it several times like that is just to emphasise that factors can be transferred at will between the term in the bracket & the exponent. That the exponential function of $x$ is identical with this limt is a standard result - showing that it is is beyond the ambit of this answer, really. Or is that actually what you're asking for?

Answer 7

Because I can :-), in R

>    whatlim <- function(n) {   
         term = sqrt( (1 + 1/n + 1/n^3)^(5*n+1) ) 
         delta = exp(2.5) - term  
         return(delta)  }

 > converge = NULL  
 > for (jj in  0:10) {  
       converge[jj+1] = whatlim(10^jj)  }

    > converge
     [1] -1.481751e+01  5.525765e-01  8.732760e-02  9.095940e-03  9.132772e-04
     [6]  9.136407e-05  9.139366e-06  8.958993e-07  2.765578e-07 -2.510824e-06
    [11] -2.519047e-06

Where pretty obviously the last two terms suffer from floating-point precision limit.

Answer 8

Let's assume the well known result $$\lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n=e\tag{1}$$ and the lemma of Thomas Andrews:

Lemma: If $\{a_n\} $ is a sequence of real or complex terms such that $n(a_n-1)\to 0$ then $a_n^n\to 1$ as $n\to \infty $.

Consider the sequence $$a_n=\dfrac{1+\dfrac{1}{n}+\dfrac {1}{n^3}}{1+\dfrac{1}{n}}$$ and check that $n(a_n-1) \to 1$ so that $a_n^n\to 1$ and using $(1)$ we get $$\lim_{n\to \infty} \left(1+\frac {1}{n}+\frac{1}{n^3}\right)^n=e\tag{2}$$ and then we get $$\lim_{n\to\infty} \left(1+\frac {1}{n}+\frac {1}{n^3}\right)^{5n+1}=e^5$$ and taking square roots we get the answer as $e^{5/2}$.