STEP III 2012 4

(i) There is quite a nice way to work out:

\displaystyle \sum_{n = 0}^\infty \frac {n^k} {n!}

for fixed k.

Since:

\displaystyle e^x = \sum_{n = 0}^\infty \frac {x^n} {n!}

We have, differentiating:

\displaystyle e^x = \sum_{n = 0}^\infty \frac {nx^{n - 1}} {n!}

So that:

\displaystyle xe^x = \sum_{n = 1}^\infty \frac {nx^n} {n!}

Setting x = 1 gives:

\displaystyle \sum_{n = 1}^\infty \frac n {n!} = e

Since we also have:

\displaystyle \sum_{n = 1}^\infty \frac 1 {n!} = e - 1

We have:

\displaystyle \sum_{n = 1}^\infty \frac {n + 1} {n!} = 2e - 1

Differentiating again we have:

\displaystyle (x + 1)e^x = \sum_{n = 1}^\infty \frac {n^2 x^{n - 1}} {n!}

So that:

\displaystyle x(x + 1)e^x = \sum_{n = 1}^\infty \frac {n^2 x^n} {n!}

and hence:

\displaystyle 2e = \sum_{n = 1}^\infty \frac {n^2 x^n} {n!}

Note that we then have:

\displaystyle \sum_{n = 1}^\infty \frac {(n + 1)^2} {n!} = \sum_{n = 1}^\infty \frac {n^2} {n!} + 2 \sum_{n = 1}^\infty \frac n {n!} + \sum_{n = 1}^\infty \frac 1 {n!} = 2e + 2e + e - 1 = 5e - 1

Differentiating again we have:

\displaystyle (2x + 1)e^x + x(x + 1)e^x = \sum_{n = 1}^\infty \frac {n^3 x^{n - 1}} {n!}

So that:

\displaystyle \sum_{n = 1}^\infty \frac {n^3} {n!} = 5e

Note finally that:

\begin{align*}\sum_{n = 1}^\infty \frac {(2n - 1)^3} {n!} & = 8 \sum_{n = 1}^\infty \frac {n^3} {n!} - 12 \sum_{n = 1}^\infty \frac {n^2} {n!} + 6 \sum_{n = 1}^\infty \frac n {n!} - \sum_{n = 1}^\infty \frac 1 {n!} \\ & = 40e - 24e + 6e - (e - 1) \\ & = 21e + 1\end{align*}

(ii)

Recall that:

\displaystyle \sum_{n = 1}^\infty \frac {x^n} n = -\ln(1 - x)

for |x| < 1.

Using partial fractions we have:

\displaystyle \frac 1 {(n + 1)(n + 2)} = \frac 1 {n + 1} - \frac 1 {n + 2}

So:

\displaystyle \frac {n^2 + 1} {2^n (n + 1)(n+2)} = \frac {n^2 + 1} {2^n (n + 1)} - \frac {n^2 + 1} {2^n (n + 2)}

For the first sum we have:

\begin{align*}\sum_{n = 0}^\infty \frac {n^2 + 1} {2^n (n + 1)} & = \sum_{n = 1}^\infty \frac {(n - 1)^2 + 1} {2^{n - 1} n} \\ & = \sum_{n = 1}^\infty \frac {n^2 - 2n + 2} {2^{n - 1} n} \\ & = \sum_{n = 1}^\infty \frac n {2^{n - 1}} - 2 \sum_{n = 1}^\infty \frac 1 {2^{n - 1}} + 4 \sum_{n = 1}^\infty \frac 1 {2^n n}\end{align*}

The second sum is a geometric series equal to \dfrac 1 {1 - \frac 1 2} = 2. The third sum is equal to \displaystyle - \ln \left(\frac 1 2\right) = \ln 2 (substitute x = \dfrac 1 2 into the series given earlier)

Noting that:

\displaystyle \sum_{n = 0}^\infty x^n = \frac 1 {1 - x}

The first sum is the derivative of (1 - x)^{-1} evaluated at x = \dfrac 1 2. That is:

\displaystyle \sum_{n = 1}^\infty \frac n {2^{n - 1}} = \frac 1 {(1 - \frac 1 2)^2} = 4

So overall:

\displaystyle \sum_{n = 0}^\infty \frac {n^2 + 1} {2^n (n + 1)} = 4 - 2 \times 4 + 4 \ln 2 = 4 \ln 2

We also have:

\begin{align*}\sum_{n = 0}^\infty \frac {n^2 + 1} {2^n (n + 2)} & = \sum_{n = 2}^\infty \frac {(n - 2)^2 + 1} {2^{n - 2} n} \\ & = \sum_{n = 1}^\infty \frac {(n - 2)^2 + 1} {2^n n} - \frac {1 + 1} {\frac 1 2} \\ & = 4 \sum_{n = 1}^\infty \frac {n^2 - 4n + 5} {2^n n} - 4 \\ & = 4 \sum_{n = 1}^\infty \frac n {2^n} - 16 \sum_{n = 1}^\infty \frac 1 {2^n} + 20 \sum_{n = 1}^\infty \frac 1 {2^n n} - 4 \end{align*}

From the previous this can be written:

\begin{align*}2 \sum_{n = 1}^\infty \frac n {2^{n - 1}} - 8 \sum_{n = 1}^\infty \frac 1 {2^{n - 1}} + 20 \sum_{n = 1}^\infty \frac 1 {2^n n} - 4 & = 2 \times 4 - 8 \times 2 + 5 \ln 2 - 4 \\ & = 20 \ln 2 - 12\end{align*}

We therefore have:

\displaystyle \sum_{n = 0}^\infty \frac {(n^2 + 1) 2^{-n}} {(n + 1)(n + 2)} = 4 \ln 2 - (20 \ln 2 - 12) = -16\ln 2 + 12

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