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 (Wink)

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- Nov 26th 2009, 09:26 PMXingyuanWho come to beat this ：）
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 (Wink) - Nov 26th 2009, 09:33 PMRockHard
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? - Nov 26th 2009, 09:49 PMDefunkt
$\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. - Nov 26th 2009, 10:01 PMRockHard
Defunkt, would the way I TRIED, lol, to show this be applicable? more specifically the last bit of my reply

- Nov 26th 2009, 10:05 PMDefunkt
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. - Nov 26th 2009, 10:10 PMRockHard
Thanks for the tips, :p 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. - Nov 26th 2009, 10:16 PMDefunkt
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. - Nov 26th 2009, 10:23 PMRockHard
Thanks for the confirmation, so this is best shown using the property of adding two series together will only converge if both series are convergent, correct?

- Nov 26th 2009, 10:42 PMXingyuan
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:

$\displaystyle \sum_{n=1}^{\infty}\left(1+\frac{1}{n}\right)a_{n} =\sum_{n=1}^{\infty}a_{n}+\sum_{n=1}^{\infty}\frac {a_{n}}{n}$

?

(Headbang) - Nov 26th 2009, 10:52 PMShanks
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 $\displaystyle +\infty(\text{or}-\infty)$? - Nov 26th 2009, 10:55 PMRockHard
Well, I think that if the way Defunkt looked at the function $\displaystyle \sum_{n=1}^{\infty}(1+\frac{1}{n})a_n$ as a sum of two series, and the property is for that too be convergent both series he broke apart must be convergent

- Nov 26th 2009, 10:58 PMShanks
maybe, we can think in a inverse direction, maybe it is easier to prove:

If $\displaystyle \sum_{n=1}^{\infty} (1+\frac{1}{n})a_{n}$ converges,

then $\displaystyle \sum_{n=1}^{\infty} a_{n}$ converges. - Nov 26th 2009, 11:15 PMDefunkt
Err.. yes, my mistake. It would probably be easier to prove noting that

1) $\displaystyle \sum_{n=1}^{\infty} a_{n}$ diverges

2) $\displaystyle \left( 1 + \frac{1}{n} \right)a_{n} \geq a_n \ \forall n \in \mathbb{N}$ - Nov 26th 2009, 11:39 PMXingyuan
- Nov 27th 2009, 12:55 AMDefunkt
lol, sorry again. shouldn't post so early in the morning. the equality will hold in absolute value, and that should work... off i go now :S