# Convergence demonstration?

I got the following assigned as homework: "Demonstrate that the following series are convergent:"

$$\sum_{k=0}^N a^k\\ \sum_{k=0}^\infty a^k \\ \sum_{k=0}^\infty ka^k \\\sum_{k=0}^\infty k(k-1)a^k\\ \sum_{k=0}^\infty k^2a^k$$

I know most of these converge when $|a| < 1$, but I'm not sure how I'm supposed to prove this, cause I don't think the ratio test applies with all the series? Especially the first one.

I apologize if this a really stupid question, cause it feels like one, but this is the first time I'm dealing with the convergence subject.

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The first one is actually a finite sum, no? So, surely that converges. – Jake Aug 27 '12 at 1:36
I think the ratio test applies here. – Joel Cohen Aug 27 '12 at 1:40
The ratio test applies for all of the series except the first one. Do you need to see how the ratio test works? I am confused by your statement after the semicolon. – Kris Williams Aug 27 '12 at 1:41
@DrKW sorry, that was supposed to be a comma. And yes, I'd appreciate an explanation of how the ratio test works. – Ana Ameer Aug 27 '12 at 1:44

The first sum converges because it contains finitely many finite terms. You can use the ratio test for proving the convergence of the others as follows:

\begin{align} \lim_{k \rightarrow \infty} {\Big|}\frac{a^{k+1}}{a^k}{\Big|} &= \lim_{k \rightarrow \infty} |a| < 1 \\ \lim_{k \rightarrow \infty} {\Big|}\frac{(k+1)a^{k+1}}{ka^k}{\Big|} &= \lim_{k \rightarrow \infty} {\Big|}{\Big(}\frac{1}{k}+1{\Big)}a{\Big|} = |a| < 1 \\ \lim_{k \rightarrow \infty} {\Big|}\frac{(k+1)ka^{k+1}}{k(k-1)a^k}{\Big|} &= \lim_{k \rightarrow \infty} {\Big|}{\Big(}\frac{1+\frac{1}{k}}{1-\frac{1}{k}}{\Big)}a{\Big|} = |a| < 1 \\ \lim_{k \rightarrow \infty} {\Big|}\frac{(k+1)^2a^{k+1}}{k^2a^k}{\Big|} &= \lim_{k \rightarrow \infty} {\Big|}{\Big(}1+\frac{1}{k}{\Big)}^2a{\Big|} = |a| < 1 \\ \end{align}

In all the above cases, the ratio limit is $< 1$. Hence all series converge absolutely.

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We recall the Ratio Test: Let $\sum\limits_{k=0}^{\infty} b_k$ be a series. If $\lim_{k \to \infty} \vert \dfrac{b_{k+1}}{b_k} \vert = L < 1$ then the series converges. If $L >1$, then the series diverges; if $L=1$ then we cannot say anything for certain.

I will show how this works for $\sum\limits_{k=0}^{\infty} k a^k$. To setup the test, we examine $$\lim_{k \to \infty} \vert \dfrac{b_{k+1}}{b_k} \vert = \lim_{k \to \infty} \vert \dfrac{(k+1)a^{k+1}}{k a^k} \vert$$ which we may rewrite as $$\lim_{k \to \infty} \vert \dfrac{k+1}{k} a \vert = |a|.$$

Therefore, the series converges whenever $|a| < 1$.

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thank you! although, from where do we get that $|a|<1$? I mean, $|a|$ could just as easily have a value bigger than $1$; or are both cases possible? that is, as per the ratio test, the series will converge if $|a|<1$ and it will diverge if $|a|>1$? – Ana Ameer Aug 27 '12 at 4:38
Right. The series converges if $|a| < 1$ and diverges if $|a| > 1$. In case $|a| = 1$, the ratio test is inconclusive. So the series may converge or diverge. Wikipedia has 3 nice examples illustrating this: en.wikipedia.org/wiki/Ratio_test#Examples – Kartik Audhkhasi Aug 27 '12 at 5:47
It is not clear if $a$ is a real number. In this case $|a|=1$ boils down to either $a=1$ or $a=-1$, and the series are easily studied. – Siminore Aug 27 '12 at 8:25