Show that $n^2\log\left(1+\frac{1}{n}\right)$ does not converge to $1$

Question

How to show that $n^2\log\left(1+\dfrac{1}{n}\right)\to 1$ is false?

I have to show that $\left(1+\dfrac{1}{n}\right)^{n^2}$ doesn't tend to $e.$

Expand $(1+1/n)^n$ and drop all terms except the two lowest order terms. You'll see that $(1+1/n)^n$ is no smaller than $2 $. — David Mitra, May 29 '13 at 17:12
It is somehow easier to prove the second one first and to deduce the first one from it. — Julien, May 29 '13 at 17:32

score 12 · Accepted Answer · answered May 29 '13 at 16:50

12

$$\log\left(1+\frac1n\right)=-\log\left(1-\frac1{n+1}\right)\geqslant\frac1{n+1}$$

answered May 29 '13 at 16:50

Did

279,727

11

I liked it better without the middle step. Thinking of $\log n$ as the area under $y=1/x$ from $1$ to $n$, we see that $$\log\left(\frac{n+1}n\right) =\int_n^{n+1}\frac{dx}x\ge \frac1{n+1},.$$ – Ted Shifrin May 29 '13 at 16:55
2

@TedShifrin I think I agree with your comment. :-)) – Did May 29 '13 at 16:56

score 8 · Answer 2 · edited May 29 '13 at 17:17

8

By Bernoulli's inequality

$$\left(1+\dfrac{1}{n}\right)^{n^2} > 1+n^2\frac{1}{n}=1+n$$

edited May 29 '13 at 17:17

Pedro

122,002

answered May 29 '13 at 17:09

N. S.

132,525

2

+1. As you can see from my deleted answer, that's the way I prefer. Maybe if you simply invoke the binomial theorem, it will sell better. – Julien May 29 '13 at 17:31
Bernoulli's inequality is a more elementary result than the binomial theorem. It just needs simple induction. – marty cohen May 29 '13 at 23:09

score 7 · Answer 3 · edited May 29 '13 at 17:08

7

$$ \log( (1 + 1/n)^n ) \to \log(e) = 1$$ So

$$ n^2 \log( 1 + 1/n) = n \log((1 + 1/n)^n) \to \infty$$

edited May 29 '13 at 17:08

Jonas Meyer

53,602

answered May 29 '13 at 16:55

Bunder

2,423

1

I think this is the slickest one. – vadim123 May 29 '13 at 17:21

Mhenni Benghorbal · Answer 4 · 2013-05-29T17:11:29.723

Hint: Use Taylor series of $\ln(1+x)$ at the point $x=0$ and see what you get. Here is the Taylor series

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

I think it is clear now.

Added:

$$\ln( 1+\frac{1}{n} )= \frac{1}{n}-{\frac {1}{2}}{\frac{1}{n^2}}+{\frac {1}{2}}{\frac{1}{n^2}}-\dots$$

$$ \implies n^2\ln( 1+\frac{1}{n} )= {n}-{\frac {1}{2}}{}+{\frac {1}{3}}{\frac{1}{n}}-\dots$$

$$ \implies \lim_{n\to \infty} n^2\ln( 1+\frac{1}{n} ) = \infty .$$

score 2 · Answer 5 · answered May 29 '13 at 23:05

2

$$n^{2}\log \left( 1+\frac{1}{n}\right) \sim n.$$

answered May 29 '13 at 23:05

Américo Tavares

38,602

score 2 · Answer 6 · answered May 29 '13 at 23:35

Another approach. Since

$$\left(1+\frac1n\right)^n\xrightarrow [n\to\infty]{}e>2.5$$

there exists $\,M\in\Bbb N\,$ s.t.

$$n>M\;\implies\; \left(1+\frac1n\right)^n>2$$

and thus, for $\,n>M$ :

$$\left(1+\frac1n\right)^{n^2}=\left[\left(1+\frac1n\right)^n\right]^n>2^n\xrightarrow [n\to\infty]{}\infty$$

score 1 · Answer 7 · answered May 29 '13 at 17:35

$n^2\log\left(1+\dfrac{1}{n}\right)$ tend to " $\infty \cdot 0$" as $n \rightarrow \infty$.

$\frac{\log\left(1+\dfrac{1}{n}\right)}{1/n^2}$ tend to "$\frac{0}{0}$" as $n \rightarrow \infty$.

By L'hopital's rule, the limit is equivalent to the limit of,

$\frac{\left( \frac{-1/n^2}{1+\frac{1}{n}} \right)}{\frac{-2}{n^3}} = \left( \frac{n}{2} \right) \left( \frac{1}{1+\frac{1}{n}} \right)$, which is "$\infty \cdot 1$" $= \infty$.

Show that $n^2\log\left(1+\frac{1}{n}\right)$ does not converge to $1$

7 Answers7