Checking series for convergence

[Lambda96]

Homework Statement

Check whether the series $\sum\limits_{k=1}^{\infty} \frac{\Bigl( 1 + \frac{1}{k} \Bigr)^{k^2}}{3^{k}}$ converges

Relevant Equations

All convergence criteria are allowed

Hi,

I am having problems with task d)

I now wanted to check the convergence using the quotient test, so $\lim_{n\to\infty} |\frac{a_{n+1}}{a_n}| < 1$

I have now proceeded as follows:

$\frac{a_{n+1}}{a_n}=\frac{\Bigl( 1 + \frac{1}{k+1} \Bigr)^{(k+1)^2}}{3^{k+1}} \cdot \frac{3^{k}}{\Bigl( 1 + \frac{1}{k} \Bigr)^{k^2}}$

$\frac{a_{n+1}}{a_n}=\frac{1}{3} \frac{\Bigl( 1 + \frac{1}{k+1} \Bigr)^{(k+1)^2}}{\Bigl( 1 + \frac{1}{k} \Bigr)^{k^2}}$

$\frac{a_{n+1}}{a_n}=\frac{1}{3} \Bigl( 1 + \frac{1}{k+1} \Bigr)^{(k+1)^2} \cdot \Bigl( 1 + \frac{1}{k} \Bigr)^{-k^2}$

Unfortunately I can't get any further now, if I form the limit $\lim_{k\to\infty} \frac{1}{3} \Bigl( 1 + \frac{1}{k+1} \Bigr)^{(k+1)^2} \cdot \Bigl( 1 + \frac{1}{k} \Bigr)^{-k^2}$ with Mathematica, $\frac{e}{3}$ must come out, unfortunately I don't know how I can show this with my expression, or should I have used a different criterion for the task?

 

[pasmith]

Look at <br /> \lim_{k \to \infty} \left|\frac{a_{k+1}}{a_k}\right| = \exp\left( \lim_{k \to \infty} (\ln |a_{k+1}| - \ln |a_{k}|) \right) and use <br /> \ln(1 + x) = x - \frac12 x^2 + O(x^3), \quad |x| &lt; 1

 

[WWGD]

Homework Statement: Check whether the series $\sum\limits_{k=1}^{\infty} \frac{\Bigl( 1 + \frac{1}{k} \Bigr)^{k^2}}{3^{k}}$ converges
Relevant Equations: All convergence criteria are allowed

Hi,

I am having problems with task d)

View attachment 336770

I now wanted to check the convergence using the quotient test, so $\lim_{n\to\infty} |\frac{a_{n+1}}{a_n}| < 1$

I have now proceeded as follows:

$\frac{a_{n+1}}{a_n}=\frac{\Bigl( 1 + \frac{1}{k+1} \Bigr)^{(k+1)^2}}{3^{k+1}} \cdot \frac{3^{k}}{\Bigl( 1 + \frac{1}{k} \Bigr)^{k^2}}$

$\frac{a_{n+1}}{a_n}=\frac{1}{3} \frac{\Bigl( 1 + \frac{1}{k+1} \Bigr)^{(k+1)^2}}{\Bigl( 1 + \frac{1}{k} \Bigr)^{k^2}}$

$\frac{a_{n+1}}{a_n}=\frac{1}{3} \Bigl( 1 + \frac{1}{k+1} \Bigr)^{(k+1)^2} \cdot \Bigl( 1 + \frac{1}{k} \Bigr)^{-k^2}$

Unfortunately I can't get any further now, if I form the limit $\lim_{k\to\infty} \frac{1}{3} \Bigl( 1 + \frac{1}{k+1} \Bigr)^{(k+1)^2} \cdot \Bigl( 1 + \frac{1}{k} \Bigr)^{-k^2}$ with Mathematica, $\frac{e}{3}$ must come out, unfortunately I don't know how I can show this with my expression, or should I have used a different criterion for the task?

Seems you could use that
$Lim_{k\rightarrow \infty}(1+\frac{1}{k})^k =e$

 

[pasmith]

Seems you could use that
$Lim_{k\rightarrow \infty}(1+\frac{1}{k})^k =e$

How does that generalize to the case where the exponent is k^2 rather than k?

 

[Lambda96]

Thank you pasmith and WWGD for your help , I have now used the root test, which allowed me to reduce the expression to $\frac{(1+\frac{1}{k})^k}{3}$ and which corresponds to the limit $\lim_{k \to \infty}$ to $\frac{e}{3}$

 

[WWGD]

How does that generalize to the case where the exponent is k^2 rather than k?

Well, Lambda96 used it to solve his problem. I only provide hints, as per PF policy.