Show that the following series converges

[SqueeSpleen]

Homework Statement

The original problem was to find for which real numbers the series
$$
\sum_{n=1}^{\infty} 2^{n} .\left( \dfrac{ n }{ n+1 } \right)^{n} x^{n}
$$
converges.
I used root criterion, it gave me a radius of convergence equal to 1/2. For x=1/2, I showed that the sequence converges to 1/e, so the series cannot be convergent. For x=-1/2 I think it's convergent but were unable to find a concise way to prove it.

Homework Equations

None really, not that I can think off. Well, I have the following expression that I don't think it can be simplified. This is what I got when I summed the nth and (n+1)th terms.
$$
- \left( \dfrac{n}{n+1} \right)^{n}+ \left( \dfrac{ n+1 }{ n+2 } \right) ^{n+1}
$$

The Attempt at a Solution

For x=-1/2 the sequence inside the series has the same absolute value than the previous one, so in absolute value the general term approaches 1/e. I think it converges, but I'm not sure how to prove it. The expression that I found trying to group the first term with the second one, the third one with the fourth one and so, is horrible and I don't think I can work with it.
I'm sure that it's easy to prove that if an oscillating series has an associated sequence that converges to a number, and the rate of convergence is good enough, then the series converges. But I'm not sure if you need a good ratio of convergence or it's valid for any convergence.
Is there any example of an oscillating sequence that converges to a real number in absolute value and it's sum is not convergent?

Anyway, I would like to see if there's an elementary way to solve this exercise because the expressions I found were horrible enough to make me think about proving a lemma rather than working with this series in particular.

Thanks for your help.

 

[ehild]

Homework Statement

The original problem was to find for which real numbers the series
$$
\sum_{n=1}^{\infty} 2^{n} .\left( \dfrac{ n }{ n+1 } \right)^{n} x^{n}
$$
converges.
I used ratio criterion, it gave me a radius of convergence equal to 1/2. For x=1/2, I showed that the sequence converges to 1/e, so the series cannot be convergent. For x=-1/2 I think it's convergent but were unable to find a concise way to prove it.

According to the ratio test a series converges if a_n+1/a_n-->q <1 when n-->infinity. Also, a_n should go to zero. What can be the value of x so as the series converge?

 

[SqueeSpleen]

Well, I know that the series converges for all x with |x|<1/2, so for all x in (-1/2,1/2). But I'm trying to check if it converges for x=-1/2, for x=1/2 I showed it doesn't converge (it goes to infinity).
Oh sorry, I didn't really used ratio test, I used root test, English is not my native language and I mixed them up.

 

[ehild]

Well, I know that the series converges for all x with |x|<1/2, so for all x in (-1/2,1/2). But I'm trying to check if it converges for x=-1/2, for x=1/2 I showed it doesn't converge (it goes to infinity).
Oh sorry, I didn't really used ratio test, I used root test, English is not my native language and I mixed them up.

A series is convergent if the sequence of its partial sums S₁, S₂ ...S_n tends to a limit L; that means that the partial sums become closer and closer to a given number L when the number of their terms increases.
if x=-1/2 you have the series Σ[(-n/(n+1)]ⁿ
Imagine that a the partial sum S_k is closer to L than a small quantity ε: L-ε<S_k < L+ε . Add the next term of the sequence to this partial sum. The magnitude of the terms of the sequence tend to e. So the next partial sum will differ from the previous one by almost e or -e. Does the series converge then?

 

[SqueeSpleen]

Oh this is really embarrasing. Sorry hahaha.
Well, I will phrase what confused me: I'm pretty sure this series has two convergent subsequences. But yes, the series per-se is not convergent as for that the term inside MUST go to zero.
This is a good example of why it's better to post the original problem instead of the one, one wants to solve hehe.
Moral of the history: Next time you put a sequence in the calculator don't only compute the partial sums for n=100, 200 and 1000, try consecutive terms too <.<

Thank you!

 

[Ray Vickson]

Homework Statement

The original problem was to find for which real numbers the series
$$
\sum_{n=1}^{\infty} 2^{n} .\left( \dfrac{ n }{ n+1 } \right)^{n} x^{n}
$$
converges.
I used root criterion, it gave me a radius of convergence equal to 1/2. For x=1/2, I showed that the sequence converges to 1/e, so the series cannot be convergent. For x=-1/2 I think it's convergent but were unable to find a concise way to prove it.

Homework Equations

None really, not that I can think off. Well, I have the following expression that I don't think it can be simplified. This is what I got when I summed the nth and (n+1)th terms.
$$
- \left( \dfrac{n}{n+1} \right)^{n}+ \left( \dfrac{ n+1 }{ n+2 } \right) ^{n+1}
$$

The Attempt at a Solution

For x=-1/2 the sequence inside the series has the same absolute value than the previous one, so in absolute value the general term approaches 1/e. I think it converges, but I'm not sure how to prove it. The expression that I found trying to group the first term with the second one, the third one with the fourth one and so, is horrible and I don't think I can work with it.
I'm sure that it's easy to prove that if an oscillating series has an associated sequence that converges to a number, and the rate of convergence is good enough, then the series converges. But I'm not sure if you need a good ratio of convergence or it's valid for any convergence.
Is there any example of an oscillating sequence that converges to a real number in absolute value and it's sum is not convergent?

Anyway, I would like to see if there's an elementary way to solve this exercise because the expressions I found were horrible enough to make me think about proving a lemma rather than working with this series in particular.

Thanks for your help.

Note that for large $n$ we have $\left(\frac{n+1}{n}\right)^n = \left( 1 + \frac{1}{n} \right)^n \sim e,$ so the $n$th term of the series looks like $e(2x)^n$ for large $n$. So, for $x = -1/2$ we have, essentially, the series $1 -1 +1 -1 + 1 -1 + \cdots$. Do you think that converges?

 

[SqueeSpleen]

Note that for large $n$ we have $\left(\frac{n+1}{n}\right)^n = \left( 1 + \frac{1}{n} \right)^n \sim e,$ so the $n$th term of the series looks like $e(2x)^n$ for large $n$. So, for $x = -1/2$ we have, essentially, the series $1 -1 +1 -1 + 1 -1 + \cdots$. Do you think that converges?

Yes. Well, in fact it does look like 1/e because the +1 is in the denominator. But the idea stands exactly as it stands with it converging to e.

That's exactly how I proved that the series doesn't converge for x=1/2, I proved that the general term converges to 1/e so the series cannot be convergent.
And that's why I'm embarrased as well viewing that seeing that it doesn't converge for x=-1/2 was completely trivial from this point. But I was too hasty to "simplify" the problem and ended up with a subsequence of the partial sums, one that converges and has an horrible expression of general terms.
Thank you for your help.