# Thread: Hard one.

1. ## Hard one.

hi

Test the series for convergence:
$\sum_{n=1}^{\infty} \left( 1 - \sqrt[n]{n} \right)$

am sure, it will be tested by using one of the comparison tests
but i can't handle it

2. Originally Posted by Miss
hi

Test the series for convergence:
$\sum_{n=1}^{\infty} \left( 1 - \sqrt[n]{n} \right)$

am sure, it will be tested by using one of the comparison tests
but i can't handle it
nvm. I thought it diverged by the nth-Term test because $n^{\frac{1}{n}}$ goes to 0, but it doesn't. It goes to 1; thus, 1 - 1 = 0. It might converge. :S

3. Originally Posted by lilaziz1
It diverges by the nth-Term test.

$\sqrt[n]{n}$ is the same as $n^{\frac{1}{n}}$
But $1 - \sqrt[n]{n} \rightarrow 0$ as $n \rightarrow \infty$

the nth term test failed here ..

4. Originally Posted by Miss
hi

Test the series for convergence:
$\sum_{n=1}^{\infty} \left( 1 - \sqrt[n]{n} \right)$

am sure, it will be tested by using one of the comparison tests
but i can't handle it
I think I have an answer that works, but I'm not 100% sure.

We know that $\sum_{n=1}^{\infty} \left( \frac{1}{n} \right)$ diverges.

I believe that, for n > 2, it can be shown that $\sqrt[n]{n} - 1 > \frac{1}{n}$. Thus $\sum_{n=1}^{\infty} \left( 1 - \sqrt[n]{n} \right)$ diverges.

I tried it as a guess, and so I worked backwards from the desired result:

$\sqrt[n]{n} - 1 > \frac{1}{n}$

$\sqrt[n]{n} > 1 + \frac{1}{n}$

$\sqrt[n]{n} > \frac{n+1}{n}$

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

$n > \frac{(n+1)^n}{n^n}$

$n^{n+1} > (n+1)^n$

It seems like the last statement holds true for all n > 2, and the steps could simply be written in reverse order to prove what we wanted to prove.

5. Originally Posted by undefined
I believe that, for n > 2, it can be shown that $\sqrt[n]{n} - 1 > \frac{1}{n}$. Thus $\sum_{n=1}^{\infty} \left( 1 - \sqrt[n]{n} \right)$ diverges.

I tried it as a guess, and so I worked backwards from the desired result:

$\sqrt[n]{n} - 1 > \frac{1}{n}$

$\sqrt[n]{n} > 1 + \frac{1}{n}$

$\sqrt[n]{n} > \frac{n+1}{n}$

$n > \left( \frac{n+1}{n} \right)^n$
To prove this last inequality, you can say that the right hand side $\left(1+\frac 1n\right)^n$ converges toward $e$. In particular, it is bounded, and hence less than $n$ for large $n$.

Alternatively, from the beginning, one can write $\sqrt[n]{n}-1=e^{\frac{\log n}{n}}-1\sim\frac{\log n}{n}$ so if you know the comparison test using asymptotic equivalence, you can conclude using $\frac{\log n}{n}\geq \frac{1}{n}$ and the divergence of $\sum_n \frac{1}{n}$.

6. Originally Posted by Laurent
To prove this last inequality, you can say that the right hand side $\left(1+\frac 1n\right)^n$ converges toward $e$. In particular, it is bounded, and hence less than $n$ for large $n$.

Alternatively, from the beginning, one can write $\sqrt[n]{n}-1=e^{\frac{\log n}{n}}-1\sim\frac{\log n}{n}$ so if you know the comparison test using asymptotic equivalence, you can conclude using $\frac{\log n}{n}\geq \frac{1}{n}$ and the divergence of $\sum_n \frac{1}{n}$.
Thanks for your explanation. I should have noticed the connection with $e$!

Regarding the asymptotic solution, would you mind explaining how you got $e^{\frac{\log n}{n}}-1\sim\frac{\log n}{n}$ ? I can verify that the LHS divided by the RHS tends to 1 as n tends to infinity, but I don't know why.

7. Originally Posted by undefined
Regarding the asymptotic solution, would you mind explaining how you got $e^{\frac{\log n}{n}}-1\sim\frac{\log n}{n}$ ? I can verify that the LHS divided by the RHS tends to 1 as n tends to infinity, but I don't know why.
You agree that $e^u-1\sim u$ as $u\to0$ (just another way of writing $\frac{e^u-1}{u}\to 1$, which is the derivative $\exp'(0)=1$). Furthermore, $\frac{\log n}{n}\to 0$ as $n\to\infty$. We may therefore compose limits and deduce $e^{\frac{\log n}{n}}-1\sim\frac{\log n}{n}$ as $n\to\infty$. Here it is!

8. Originally Posted by Laurent
You agree that $e^u-1\sim u$ as $u\to0$ (just another way of writing $\frac{e^u-1}{u}\to 1$, which is the derivative $\exp'(0)=1$). Furthermore, $\frac{\log n}{n}\to 0$ as $n\to\infty$. We may therefore compose limits and deduce $e^{\frac{\log n}{n}}-1\sim\frac{\log n}{n}$ as $n\to\infty$. Here it is!
Thank you! It's very clear now.