Evaluate the limit of this harmonic series as n tends to infinity

[Aurelius120]

Homework Statement

Evaluate $$\lim_{n\rightarrow \infty}\left(\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+........\frac{1}{n}}{n^2}\right)^n$$

Relevant Equations

$$\lim_{x \to{+}\infty}{f(x)^{g(x)}}=e^{\left(\lim_{x \to{+}\infty}{(f(x)-1)\cdot g(X)}\right)}$$ $$\text{if}\lim_{n\rightarrow \infty}f(x)=1\text{ and } \lim_{n\rightarrow \infty}g(x)=\infty$$

To use the formula above, I have to prove that $$\lim_{n\rightarrow \infty}f(x)=\lim_{n\rightarrow \infty}\left(\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+........\frac{1}{n}}{n^2}\right)=1$$
To prove so, I tried using L'Hopital's Rule:
$$\lim_{n\rightarrow \infty}f(x)=\lim_{n\rightarrow \infty}\frac{-1n^{(-2)}}{2n}$$ But this gives zero.

 

[Hill]

The numerator in the last expression is incorrect.

 

[Aurelius120]

The numerator in the last expression is incorrect.

Why so ?
$$\frac{d\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+..........\frac{1}{n}\right)}{dn}=\frac{dn^{-1}}{dn}=-n^{-2}$$

 

[Hill]

$\frac{d\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+..........\frac{1}{n}\right)}{dn} \neq \frac{dn^{-1}}{dn}$

Rather,
$\frac{d\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{n}\right)}{dn}=\frac{dn^{-1}}{dn}$

 

[Aurelius120]

$\frac{d\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+..........\frac{1}{n}\right)}{dn} \neq \frac{dn^{-1}}{dn}$

Rather,
$\frac{d\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{n}\right)}{dn}=\frac{dn^{-1}}{dn}$

Why is that ? Since the constants after that will also be zero?
Also even we go the other way:
$$\lim_{n\rightarrow \infty}f(x)=\lim_{n\rightarrow \infty}\left(\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+........\frac{1}{n}}{n^2}\right)=\lim_{n\rightarrow \infty}\left(\frac{1}{n^2}+\frac{1}{2n^2}+\frac{1}{3n^2}+\frac{1}{4n^2}+........\frac{1}{n^3}\right)$$ which is still zero.

 

[PeroK]

I don't see the point of using L'Hopital. The answer should be clear using a crude estimate for the partial sums of the harmonic series.

 

[PeroK]

PS just try $n = 2, 3 \dots$, to see what's happening:
$$a(2) = \bigg (\frac{1 + \frac 1 2}{4}\bigg)^2 = \bigg(\frac{3}{8}\bigg)^2$$
$$a(3) = \bigg (\frac{1 + \frac 1 2 + \frac 1 3}{9}\bigg)^3 = \bigg(\frac{11}{54}\bigg)^3$$That sequence definitely does not converge to $1$.

 

[PeroK]

I have to prove that $$\lim_{n\rightarrow \infty}f(x)=\lim_{n\rightarrow \infty}\left(\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+........\frac{1}{n}}{n^2}\right)=1$$

Good luck with proving that!

 

[Hill]

Why is that ? Since the constants after that will also be zero?

Because $$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+........\frac{1}{n} \neq const.+\frac{1}{n}$$
Actually,
$$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+........\frac{1}{n} = 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+........+\frac{1}{n-3}+\frac{1}{n-2}+\frac{1}{n-1}+\frac{1}{n}$$

 

[Aurelius120]

Well the solution involves somehow using the "relevant equations" formula. For which
$$\lim_{n\rightarrow \infty}f(x)=1$$ but
$$\lim_{n\rightarrow \infty}f(x)=\lim_{n\rightarrow \infty}\left(\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+........\frac{1}{n}}{n^2}\right)$$$$=\lim_{n\rightarrow \infty}\left(\frac{1}{n^2}+\frac{1}{2n^2}+\frac{1}{3n^2}+\frac{1}{4n^2}+........\frac{1}{n^3}\right)=0$$

Good luck with proving that!

So that is a problem

 

[PeroK]

It's not clear to me why you're rejecting what appears to be the correct answer. Should the original problem involve an nth root, rather than a power of n?

 

[Aurelius120]

The question in the book is

The solution uses $$\lim_{n\rightarrow \infty}e^{\left(\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+........\frac{1}{n}}{n^2}\right)\cdot n}$$
I figured the solution evaluated the part after 1 instead of the entire limit

It's not clear to me why you're rejecting what appears to be the correct answer. Should the original problem involve an nth root, rather than a power of n?

That is why I wanted to show that limit was 1

 

[WWGD]

If you're faced with this question in an exam, just answer 1. It's the only possible choice. The expression in the parenthesis is non-negative, so the only option is 1, meaning the parenthesis must equal 0.

 

[mathhabibi]

The limit is $$\lim_{n\rightarrow\infty}\left(\frac{H_n}{n^2}\right)^n=\lim_{n\rightarrow\infty}\frac{H^n_n}{n^{2n}}$$Where $H_n$ is the $n$th harmonic number. The numerator grows much slower than the denominator and so the answer is just $0$.

 

[Aurelius120]

The limit is $$\lim_{n\rightarrow\infty}\left(\frac{H_n}{n^2}\right)^n=\lim_{n\rightarrow\infty}\frac{H^n_n}{n^{2n}}$$Where $H_n$ is the $n$th harmonic number. The numerator grows much slower than the denominator and so the answer is just $0$.

This is the given solution though

 

[mathhabibi]

View attachment 340474
This is the given solution though

Take the natural log of both sides. Also this isn't the original limit as in the question.

 

[WWGD]

I think ln(n) is an upper bound for the sum $\Sigma_{i=1}^n \frac{1}{i}$. If so, the power is bound by $\frac{ln(n)}{n}$, which goes to $0$ as $n$ goes to $\infty$. (i.e., using that $ln(e^n=n)$I experimented with powers of $e; e^n; n=1,2,...$ and this bound seemed to hold up. Obviously both sequences diverge, but $ln(n)$ does so much faster than the Harmonic series.

 

[mathhabibi]

I think ln(n) is an upper bound for the sum $\Sigma_{i=1}^n \frac{1}{i}$.

No. It's $\ln(n)+\gamma$ (where $\gamma$ is the Euler-Mascheroni constant). However, $\ln n$ is pretty close to the harmonic numbers so replacing the numbers with that function shouldn't change anything.