# Find all real numbers such that the series converges!

• February 20th 2014, 11:42 PM
mynameisanthonyg2013
Find 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.
• February 20th 2014, 11:57 PM
romsek
Re: Find all real numbers such that the series converges!
deleted
• February 21st 2014, 12:06 AM
mynameisanthonyg2013
Re: 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?
• February 21st 2014, 12:13 AM
romsek
Re: Find all real numbers such that the series converges!
yeah ok, let me think about it some.
• February 21st 2014, 12:13 AM
mynameisanthonyg2013
Re: Find all real numbers such that the series converges!
Sure thanks!
• February 21st 2014, 12:17 AM
mynameisanthonyg2013
Re: 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!
• February 21st 2014, 04:44 AM
Plato
Re: Find all real numbers such that the series converges!
Quote:

Originally Posted by mynameisanthonyg2013
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?

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 $\sum\dfrac{c}{n}$.
• February 21st 2014, 11:30 AM
romsek
Re: Find all real numbers such that the series converges!
Quote:

So use the compassion test with $\sum \frac{c}{n}$.
do you mean comparison test?