Choosing a test for infinite series.

[Drezzan]

Homework Statement

The problem is part of a review and we are only to determine if the series converges or diverges by any test, and state the test.

$\sum_{n=1}^\infty(\frac{k}{k+1})^k$


My work so far

I know that the root test gives an inconclusive answer and from there I moved to looking at the series like a geometric. I have no idea if this actually works and could use input.
$$\sum_{n=1}^\infty(\frac{k}{k+1})^k$$ geometric series $r=\frac{k}{k+1}$

$$\lim_{k\rightarrow\infty}{\frac{k}{k+1}}=1$$ r>0 series converges.


Does this work ? I keep second guessing myself but I can't find another test that works.

 

[Dick]

Homework Statement

The problem is part of a review and we are only to determine if the series converges or diverges by any test, and state the test.

$\sum_{n=1}^\infty(\frac{k}{k+1})^k$


My work so far

I know that the root test gives an inconclusive answer and from there I moved to looking at the series like a geometric. I have no idea if this actually works and could use input.
$$\sum_{n=1}^\infty(\frac{k}{k+1})^k$$ geometric series $r=\frac{k}{k+1}$

$$\lim_{k\rightarrow\infty}{\frac{k}{k+1}}=1$$ r>0 series converges.


Does this work ? I keep second guessing myself but I can't find another test that works.

The first check you should apply to a series is to check whether the terms of your series approach 0 as k->infinity. If not it diverges. Do they?

 

[Drezzan]

True! I would like to point out I read the formula wrong and its if r> or =1 then it diverges.

Taking a look at the problem again, if we take the highest power in the numerator and the denominator, like in my last step, we get it going to 1. That means it does diverge by the divergence test.

My question is since it is $(k+1)^k$ in the denominator wouldn't that outpace the k in the numerator? I know I can factor a k out of the bottom to get $\frac{1}{1+\frac{1}{k}}$ but I have only employed this method after evaluating something like the ratio test. Is it still valid?

 

[Dick]

True! I would like to point out I read the formula wrong and its if r> or =1 then it diverges.

Taking a look at the problem again, if we take the highest power in the numerator and the denominator, like in my last step, we get it going to 1. That means it does diverge by the divergence test.

My question is since it is $(k+1)^k$ in the denominator wouldn't that outpace the k in the numerator? I know I can factor a k out of the bottom to get $\frac{1}{1+\frac{1}{k}}$ but I have only employed this method after evaluating something like the ratio test. Is it still valid?

It's not really obvious what its limiting behaviour is. Would you recognize what the limit of (1+1/k)^k is? Maybe it was mentioned in your text? It's pretty closely related to the limit of your sequence.

 

[lurflurf]

$$\lim_{k\rightarrow\infty} \left(\frac{k}{k+1}\right)^k=\left(\lim_{k \rightarrow \infty} \left(1+\frac{1}{k}\right)^k\right)^{-1}$$

This is a well known (nonzero) limit so the series diverges by the limit test.

You were trying the comparison test, but it does not apply because we need all the terms to be less than some number less than one, and even though all terms are less than one they approach it so no number less than one exceeds them all.