If $\displaystyle \sum_{n=1}^{\infty}a_{n}$ diverges
show that:
$\displaystyle \sum_{n=1}^{\infty}\left(1+\frac{1}{n}\right)a_{n}$
diverges
I never get these general term series question, multiplied by some function
I know that simply by the divergence test or that it is the harmonic series.
$\displaystyle \sum_{n=1}^{\infty}(1+\frac{1}{n})$
$\displaystyle \lim_{n\to\infty} 1 + lim_{n\to\infty} \frac{1}{n} =1 $
which is not 0 and by the divergence test if the limit of $\displaystyle a_n \ne 0$ diverges
$\displaystyle \sum_{n=1}^{\infty}(1+\frac{1}{n})$
Could say that the An is just like multiplying a constant and which multiplying by a constant will not affect if a series diverges or convergences? just the sum?
$\displaystyle \sum_{n=1}^{\infty}(1+\frac{1}{n})a_n$
similarly
$\displaystyle \sum_{n=1}^{\infty}\frac{a_n(1+n)}{n}$
This is a Harmonic series which diverges...I am incorrect?
$\displaystyle \sum_{n=1}^{\infty}\left(1+\frac{1}{n}\right)a_{n} = $ $\displaystyle \sum_{n=1}^{\infty} a_{n} + \sum_{n=1}^{\infty}\left(\frac{1}{n}\right)a_{n}$.
What you have here is either a sum of two divergent series or a sum of a convergent and a divergent series, both of which diverge.
Well, I don't really understand what you're saying there :P Try organizing your thoughts more clearly and then presenting them.
If you are saying that multiplying the series by $\displaystyle \left( 1 + \frac{1}{n} \right)$ is like multiplying by a constant, that's wrong. It simply is not a constant. And no, it is not a harmonic series -- http://en.wikipedia.org/wiki/Harmoni...s_(mathematics) this is the only form of a harmonic series.
Thanks for the tips, I am just so eager to learn, I make a mess of myself, I was trying to say that multiplying by $\displaystyle A_n$ to the series of
$\displaystyle \sum_{n=1}^{\infty}(1+\frac{1}{n})$, you can treat $\displaystyle A_n$ as constant, more than likely this is where I am incorrect, and that multiplying a constant to a series will not affect its convergence or divergence just it's sum.
Also tried saying that
$\displaystyle \sum_{n=1}^{\infty}(1+\frac{1}{n})$
can be written as
$\displaystyle \sum_{n=1}^{\infty}\frac{n+1}{n}$
which looks to be of similar form to the harmonic series to me, but what the hell, I am more incorrect than correct, when I post messes like these, lol.
The harmonic series is ONLY this: $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}$. NOTHING else. Not everything that is divided by n is a harmonic series :P
The part about multiplying by a constant is true, though. If you multiply a series by a non-zero constant, its convergence will not change -- only the value of the series (if it converges). The thing here is that $\displaystyle 1+ \frac{1}{n}$ is not a constant (or do you mean the original divergent series $\displaystyle \sum_{n=1}^{\infty}a_n$? either way, both aren't constants), even though it goes to 1 as n goes to infinity.
If $\displaystyle \sum_{n=1}^{\infty}\left(1+\frac{1}{n}\right)a_{n}$ converges
we can decompose it into two part:
$\displaystyle \sum_{n=1}^{\infty}\left(1+\frac{1}{n}\right)a_{n} =\sum_{n=1}^{\infty}a_{n}+\sum_{n=1}^{\infty}\frac {a_{n}}{n}$
In other words, we can treat it as finite summation.
but if it diverges, I think we can't treat it as finite summation, so If
$\displaystyle \sum_{n=1}^{\infty}a_{n}$ diverges,
can we have:
$\displaystyle \sum_{n=1}^{\infty}\left(1+\frac{1}{n}\right)a_{n} =\sum_{n=1}^{\infty}a_{n}+\sum_{n=1}^{\infty}\frac {a_{n}}{n}$
?
let $\displaystyle a_n^{+}=max\{a_n,0\}, a_n^{-}=max\{-a_n,0\}$
then $\displaystyle \sum_{n=1}^{\infty} a_{n}=\sum_{n=1}^{\infty} a_{n}^{+}-\sum_{n=1}^{\infty} a_{n}^{-}$.
Since $\displaystyle \sum_{n=1}^{\infty} a_{n}$ divergens, then at least one of them diverges.
If only one of them diverges; then obviously the conclusion is true.
If both of them diverges, thennnnnnn......
what happens if $\displaystyle \sum_{n=1}^{\infty} a_{n}$ diverges to $\displaystyle +\infty(\text{or}-\infty)$?