Solving an infinite geometric series.

Question

Let $S_k, k=1,2,...,100$, denote the sum of the infinite geometric series whose first term is $\dfrac{k-1}{k!}$ and the common ratio is $\dfrac{1}{k}$. Then the value of $\dfrac{100^2}{100!}+\displaystyle\sum^{100}_{k=1}\left|\left(k^2-3k+1\right)S_k\right|$ is

My Attempt: $$S_k=\dfrac{\dfrac{k-1}{k!}}{1-\dfrac{1}{k}}=\dfrac{k}{k!}=\dfrac{1}{\left(k-1\right)!}$$ $$\displaystyle\sum^{100}_{k=1}\left|\left(k^2-3k+1\right)S_k\right|=\displaystyle\sum^{100}_{k=1}\left|\dfrac{\left(k^2-3k+1\right)}{\left(k-1\right)!}\right|=\displaystyle\sum^{100}_{k=1}\left|\dfrac{\left(k-1\right)\left(k-2\right)-1}{\left(k-1\right)!}\right|=\displaystyle\sum^{100}_{k=1}\left|\dfrac{1}{\left(k-3\right)!}-\dfrac{1}{\left(k-1\right)!}\right|$$

Substituting values of $k$ in $\displaystyle\sum^{100}_{k=1}\left|\dfrac{1}{\left(k-3\right)!}\right|$ gives factorial of negative integers like $\dfrac{1}{(-2)!}$.

How do I proceed from here?

Answer 1

We have $S_1=0$ and for $k>1$, $$S_k = \frac{k-1}{k!}\sum_{n=0}^\infty \frac{1}{k^n}=\frac{k-1}{k!}\frac{1}{1-\frac{1}{k}} = \frac{k-1}{k!}\frac{k}{k-1}=\frac{1}{(k-1)!}$$

(Notice that this last formula doesn't work for $k=1$, that case need to be treated separately).

Then, as observed in other answers, for $k>2$ we have $k^2-3k+1>0$. The first two terms of the sum are $\left|(1^2-3\cdot 1+1)\cdot S_1\right|+\left|(2^2-3\cdot 2+1)\cdot S_2\right| = \left|(-1)\cdot 0\right|+\left|(-1)\cdot \frac{1}{(2-1)!}\right| = 1$

Then

\begin{align} &\frac{100^2}{100!}+\sum^{100}_{k=1}\left|(k^2-3k+1)S_k\right|\\ & = \frac{100^2}{100!}+1+\sum^{100}_{k=3}(k^2-3k+1)S_k\\ &=\frac{100^2}{100!}+1+\sum_{k=3}^{100}\frac{(k-1)(k-2)-1}{(k-1)!}\\ &=\frac{100^2}{100!}+1+\sum_{k=3}^{100}\frac{1}{(k-3)!}-\sum_{k=3}^{100}\frac{1}{(k-1)!}\\ &=\frac{100^2}{100!}+1+\sum_{k=0}^{97}\frac{1}{k!}-\sum_{k=2}^{99}\frac{1}{k!}\\ &=\frac{100^2}{100!}+1+\frac{1}{0!}+\frac{1}{1!}-\frac{1}{98!}-\frac{1}{99!}\\ &=\frac{100^2}{100!}+3-\frac{99\cdot 100}{100!}-\frac{100}{100!}\\ &=\frac{100^2}{100!}+3-\frac{99\cdot 100 + 100}{100!}\\ &=\frac{100^2}{100!}+3-\frac{100^2}{100!}\\ &=3 \end{align}

Answer 2

Hint: Take out the first two terms as

$$ \sum^{100}_{k=1}\left|\left(k^2-3k+1\right)S_k\right| = |(1^2 - 3 + 1)S_1| + |(2^2 - 6 + 1)S_2| + \sum^{100}_{k=3}\left|\left(k^2-3k+1\right)S_k\right| $$

Now notice that $S_k \ge 0$ for all $k \ge 1$ and for $k \ge 3, k^2 - 3k + 1 > 0$, i.e. the sign will always be positive so you can get rid of the mod notation for $k \ge 3$ and your sum becomes

$$ \sum^{100}_{k=1}\left|\left(k^2-3k+1\right)S_k\right| = S_1 + S_2 + \sum^{100}_{k=3}(k^2-3k+1)S_k $$

If you do the simplifications correctly you should get

$$ \frac{100^2}{100!} + \sum^{100}_{k=1}\left|\left(k^2-3k+1\right)S_k\right| = 4 $$