Is there a real number c such that the series:

∑ (e - (1+ 1/n)^n + c/n), where the series goes from n=1 to n=∞, is convergent?

I tried to use ratio test but it turned out really messy. Any other ideas?

Thanks.

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- Feb 20th 2014, 10:42 PMmynameisanthonyg2013Find all real numbers such that the series converges!
Is there a real number c such that the series:

∑ (e - (1+ 1/n)^n + c/n), where the series goes from n=1 to n=∞, is convergent?

I tried to use ratio test but it turned out really messy. Any other ideas?

Thanks.

- Feb 20th 2014, 10:57 PMromsekRe: Find all real numbers such that the series converges!
deleted

- Feb 20th 2014, 11:06 PMmynameisanthonyg2013Re: Find all real numbers such that the series converges!
Yes I knew that and I did come to the harmonic series conclusion, but isn't it true that lim as n→∞ (1+1/n)^n is different from summation (1+1/n)^n as n from 1 to infinity?

- Feb 20th 2014, 11:13 PMromsekRe: Find all real numbers such that the series converges!
yeah ok, let me think about it some.

- Feb 20th 2014, 11:13 PMmynameisanthonyg2013Re: Find all real numbers such that the series converges!
Sure thanks!

- Feb 20th 2014, 11:17 PMmynameisanthonyg2013Re: Find all real numbers such that the series converges!
Also since you are answering all my questions, I might as well post one more for you to look at. That doesn't look too hard, but there's got to be some trick that I can't think of at the moment!

- Feb 21st 2014, 03:44 AMPlatoRe: Find all real numbers such that the series converges!
In the case that $c>0 $ then:

It is well known that ${\left( {1 + {n^{ - 1}}} \right)^n}$ is a monotone increasing sequence $\to e$.

Therefore $(\forall n)\left[e-{\left( {1 + {n^{ - 1}}} \right)^n}>0\right]$. So use the comparison test with $\sum\dfrac{c}{n}$. - Feb 21st 2014, 10:30 AMromsekRe: Find all real numbers such that the series converges!Quote:

So use the compassion test with $\displaystyle \sum \frac{c}{n}$.