Summation of a series

**AgentSmith** · January 7th 2012, 11:55 PM

Let $S_k$ , $k = 1,\ 2, \cdots ,\ 100$ , denote the sum of the infinite geometric series whose first term is $\frac{k-1}{k!}$ and common ration is $\frac{1}{k}$ then find the value of $\frac{100^2}{100!}+\sum_{k=1}^{100}|(k^2-3k+1)S_k|$ .

How do I do this?

**AgentSmith** · January 8th 2012, 12:04 AM

Originally Posted by CaptainBlack

Have you considered starting by writing $S_k$ as a function of $k$ ?
CB

Yes. I got $S_k=\frac{1}{(k-1)!}$ .

**sbhatnagar** · January 8th 2012, 12:29 AM

$S_k$ is an infinite geometric series.

1.Sum of an infinite geometric series is given by $S=\frac{a}{1-r}$

therefore $S_k= \frac{\frac{k-1}{k!}}{1-\frac{1}{k}}$

$\implies S_k = \frac{1}{(k-1)!}$

2. Note That:
$\begin{align*} \sum_{k=1}^{100}\Big| \frac{k^2-3k+1}{(k-1)!}\Big|&= \sum_{k=1}^{100}\Big| \frac{(k-1)^2-k}{(k-1)!}\Big| \\ &=\sum_{k=1}^{100}\Big| \frac{k-1}{(k-2)!}-\frac{k}{(k-1)!}\Big| \\ &= |-1|+|-1|+\Big| \frac{2}{1!}-\frac{3}{2!}\Big|+\Big| \frac{3}{2!}-\frac{4}{3!}\Big|+\cdots + \Big| \frac{99}{98!}-\frac{100}{99!}\Big| \\ &= 1+1+2-\frac{100}{99!} \\ &=4-\frac{100}{99!}\end{align*}$

**AgentSmith** · January 8th 2012, 12:42 AM

Originally Posted by sbhatnagar

$S_k$ is an infinite geometric series.

1.Sum of an infinite geometric series is given by $S=\frac{a}{1-r}$

therefore $S_k= \frac{\frac{k-1}{k!}}{1-\frac{1}{k}}$

$\implies S_k = \frac{1}{(k-1)!}$

2. Note That:
$\begin{align*} \sum_{k=1}^{100}\Big| \frac{k^2-3k+1}{(k-1)!}\Big|&= \sum_{k=1}^{100}\Big| \frac{(k-1)^2-k}{(k-1)!}\Big| \\ &=\sum_{k=1}^{100}\Big| \frac{k-1}{(k-2)!}-\frac{k}{(k-1)!}\Big| \\ &= |-1|+|-1|+\Big| \frac{2}{1!}-\frac{3}{2!}\Big|+\Big| \frac{3}{2!}-\frac{4}{3!}\Big|+\cdots + \Big| \frac{99}{98!}-\frac{100}{99!}\Big| \\ &= 1+1+2-\frac{100}{99!} \\ &=4-\frac{100}{99!}\end{align*}$

Thanks!

That should mean my final answer is $4-\frac{100}{99!}+\frac{100^2}{100!}$ , but the answer provided to me is 4.

**sbhatnagar** · January 8th 2012, 12:53 AM

Originally Posted by AgentSmith

Thanks!

That should mean my final answer is $4-\frac{100}{99!}+\frac{100^2}{100!}$ , but the answer provided to me is 4.

$\begin{align*}\frac{100^2}{100!}-\frac{100}{99!} &= \frac{100^2}{100!}-\frac{100}{99!}\cdot \frac{100}{100}\\ &=\frac{100^2}{100!}-\frac{100^2}{100!}\\&=0 \end{align*}$

**AgentSmith** · January 8th 2012, 01:48 AM

Originally Posted by sbhatnagar

$\begin{align*}\frac{100^2}{100!}-\frac{100}{99!} &= \frac{100^2}{100!}-\frac{100}{99!}\cdot \frac{100}{100}\\ &=\frac{100^2}{100!}-\frac{100^2}{100!}\\&=0 \end{align*}$

Thank you! I can't believe, I was being so stupid...

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