# Thread: Who come to beat this ：）

1. ## Who come to beat this ：）

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

show that:

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

diverges

2. 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.

$\sum_{n=1}^{\infty}(1+\frac{1}{n})$

$\lim_{n\to\infty} 1 + lim_{n\to\infty} \frac{1}{n} =1$

which is not 0 and by the divergence test if the limit of $a_n \ne 0$ diverges

$\sum_{n=1}^{\infty}(1+\frac{1}{n})$

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?

$\sum_{n=1}^{\infty}(1+\frac{1}{n})a_n$

similarly

$\sum_{n=1}^{\infty}\frac{a_n(1+n)}{n}$

This is a Harmonic series which diverges...I am incorrect?

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

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.

4. Defunkt, would the way I TRIED, lol, to show this be applicable? more specifically the last bit of my reply

5. 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 $\left( 1 + \frac{1}{n} \right)$ 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.

6. 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 $A_n$ to the series of
$\sum_{n=1}^{\infty}(1+\frac{1}{n})$, you can treat $A_n$ 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

$\sum_{n=1}^{\infty}(1+\frac{1}{n})$

can be written as

$\sum_{n=1}^{\infty}\frac{n+1}{n}$

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.

7. The harmonic series is ONLY this: $\sum_{n=1}^{\infty}\frac{1}{n}$. 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 $1+ \frac{1}{n}$ is not a constant (or do you mean the original divergent series $\sum_{n=1}^{\infty}a_n$? either way, both aren't constants), even though it goes to 1 as n goes to infinity.

8. 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?

9. Originally Posted by Defunkt
$\sum_{n=1}^{\infty}\left(1+\frac{1}{n}\right)a_{n} =$ $\sum_{n=1}^{\infty} a_{n} + \sum_{n=1}^{\infty}\left(\frac{1}{n}\right)a_{n}$.

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

we can decompose it into two part:

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

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

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

can we have:
$\sum_{n=1}^{\infty}\left(1+\frac{1}{n}\right)a_{n} =\sum_{n=1}^{\infty}a_{n}+\sum_{n=1}^{\infty}\frac {a_{n}}{n}$
?

10. let $a_n^{+}=max\{a_n,0\}, a_n^{-}=max\{-a_n,0\}$
then $\sum_{n=1}^{\infty} a_{n}=\sum_{n=1}^{\infty} a_{n}^{+}-\sum_{n=1}^{\infty} a_{n}^{-}$.
Since $\sum_{n=1}^{\infty} a_{n}$ 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 $\sum_{n=1}^{\infty} a_{n}$ diverges to $+\infty(\text{or}-\infty)$?

11. Well, I think that if the way Defunkt looked at the function $\sum_{n=1}^{\infty}(1+\frac{1}{n})a_n$ as a sum of two series, and the property is for that too be convergent both series he broke apart must be convergent

12. maybe, we can think in a inverse direction, maybe it is easier to prove:
If $\sum_{n=1}^{\infty} (1+\frac{1}{n})a_{n}$ converges,
then $\sum_{n=1}^{\infty} a_{n}$ converges.

13. Err.. yes, my mistake. It would probably be easier to prove noting that
1) $\sum_{n=1}^{\infty} a_{n}$ diverges
2) $\left( 1 + \frac{1}{n} \right)a_{n} \geq a_n \ \forall n \in \mathbb{N}$

14. Originally Posted by Defunkt
Err.. yes, my mistake. It would probably be easier to prove noting that
1) $\sum_{n=1}^{\infty} a_{n}$ diverges
2) $\left( 1 + \frac{1}{n} \right)a_{n} \geq a_n \ \forall n \in \mathbb{N}$
If $a_{n}>0$ for $\forall n \in \mathbb{N}^{+}$

then we have:
$\left( 1 + \frac{1}{n} \right)a_{n} \geq a_n \ \forall n \in \mathbb{N}$

so......I think "2)" will not correct for all $n$

15. 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

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