Convergent or divergent? That is the question

Proving the convergence of the series.

$$ $T_{n+1} = \dfrac{(n+1)^4}{2^{n+1}}$

$$ $T_{n} = \dfrac{(n)^4}{2^{n}}$

$\displaystyle \lim_{n \rightarrow \infty} \dfrac{T_{n+1}}{T_{n}} = \lim_{n \rightarrow \infty} \dfrac{(n+1)^4 \times 2^n}{n^4 \times 2^{n+1}} = \dfrac{1}{2}$

$$ Thus, by the ratio test, the series converges, and any rearrangement of terms is justified.

$$ S = $\dfrac{1}{2} + \dfrac{16}{4} + \dfrac{81}{8} + \dfrac{256}{16} + \dfrac{625}{32} .... \infty$ ..(i)

$$ $\dfrac{S}{2} = \dfrac{1}{4} + \dfrac{16}{8} + \dfrac{81}{16} + \dfrac{256}{32} .... \infty$ ..(ii)

$$ Subtracting (ii) from (i),

$\dfrac{S}{2} = \dfrac{1}{2} + \dfrac{15}{4} + \dfrac{65}{8} + \dfrac{175}{16} + \dfrac{369}{32} .... \infty$ ..(iii)

$$ $\dfrac{S}{4} = \dfrac{1}{4} + \dfrac{15}{8} + \dfrac{65}{16} + \dfrac{175}{32} + .... \infty$ ..(iv)

$$ Subtracting (iv) from (iii),

$$ $\dfrac{S}{4} = \dfrac{1}{2} + \dfrac{14}{4} + \dfrac{50}{8} + \dfrac{110}{16} + \dfrac{194}{32} .... \infty$ ..(v)

$$ $\dfrac{S}{8} = \dfrac{1}{4} + \dfrac{14}{8} + \dfrac{50}{16} + \dfrac{110}{32} + .... \infty$ ..(vi)

$$ Subtracting (vi) from (v),

$$ $\dfrac{S}{8} = \dfrac{1}{2} + \dfrac{13}{4} + \dfrac{36}{8} + \dfrac{60}{16} + \dfrac{84}{32} .... \infty$ ..(vii)

$$ $\dfrac{S}{16} = \dfrac{1}{4} + \dfrac{13}{8} + \dfrac{36}{16} + \dfrac{60}{32} .... \infty$ ..(viii)

$$ Subtracting (viii) from (vii)

$$ $\dfrac{S}{16} = \dfrac{1}{2} + \dfrac{12}{4} + \dfrac{23}{8} + \dfrac{24}{16} + \dfrac{24}{32} .... \infty$

$$ The later terms form an infinite G.P with first term a = $\dfrac{24}{16}$ and common ratio r = $\dfrac{1}{2}$

$$ $\therefore \dfrac{S}{16} = \dfrac{1}{2} + \dfrac{12}{4} + \dfrac{23}{8} + \dfrac{24}{16} \times \dfrac{1}{1-\dfrac{1}{2}} = \dfrac{1}{2} + \dfrac{12}{4} + \dfrac{23}{8} + \dfrac{24}{8}$

$$ $\therefore \dfrac{S}{16} = \dfrac{75}{8}$

$$ $\therefore S = 150$

$$ $S = 150 = 5^2 \times 2 \times 3$

$$ Number of positive factors of S ( A ) = $(2+1) \times (1+1) \times (1+1) = 12$

$$ $\therefore S - A = 150 - 12 = 138$

Notice that the first term of $\displaystyle S = \sum_{n=0}^{\infty} \dfrac{n^k}{2^n}$ is equal to 0, so the sum would be the same if we started with the second term. Thus we can see that $\displaystyle \sum_{n=0}^{\infty} \dfrac{n^k}{2^n} = \sum_{n=0}^{\infty} \dfrac{(n+1)^k}{2^{n+1}} \implies 2\sum_{n=0}^{\infty} \dfrac{n^k}{2^n} = \sum_{n=0}^{\infty} \dfrac{(n+1)^k}{2^n} \implies \sum_{n=0}^{\infty} \dfrac{n^k}{2^n} = \sum_{n=0}^{\infty} \dfrac{(n+1)^k-n^k}{2^n}$ We now have a general rule that rewrites a degree k polynomial in the numerator as a degree k-1 polynomial in the numerator. $\begin{aligned} \sum_{n=0}^{\infty} \dfrac{n^4}{2^n} &&&&&&& \sum_{n=0}^{\infty} \dfrac{n^4}{2^n} = \sum_{n=0}^{\infty} \dfrac{4n^3+6n^2+4n+1}{2^n} \\ = \sum_{n=0}^{\infty} \dfrac{4n^3+6n^2+4n+1}{2^n} &&&&&&& \sum_{n=0}^{\infty} \dfrac{n^3}{2^n} = \sum_{n=0}^{\infty} \dfrac{3n^2+3n+1}{2^n} \\ = \sum_{n=0}^{\infty} \dfrac{18n^2+16n+5}{2^n} &&&&&&& \sum_{n=0}^{\infty} \dfrac{n^2}{2^n} = \sum_{n=0}^{\infty} \dfrac{2n+1}{2^n} \\ = \sum_{n=0}^{\infty} \dfrac{52n+23}{2^n} &&&&&&& \sum_{n=0}^{\infty} \dfrac{n}{2^n} = \sum_{n=0}^{\infty} \dfrac{1}{2^n} \\ = \sum_{n=0}^{\infty} \dfrac{75}{2^n} = 150 \end{aligned}$ 150 has 12 factors, so the answer is 138.

Consider $S_k = \displaystyle \sum_{n=0}^\infty \frac {n^k}{2^n}$ . By ratio text $S_k$ converges.

$\begin{aligned} S_0 & = \sum_{n=0}^\infty \frac 1{2^n} = \frac 1{1-\frac 12} = 2 \end{aligned}$

$\begin{aligned} S_1 & = \sum_{\color{#3D99F6}n=0}^\infty \frac n{2^n} = \sum_{\color{#D61F06}n=1}^\infty \frac n{2^n} = \sum_{\color{#3D99F6}n=0}^\infty \frac {n+1}{2^{n+1}} = \frac 12 \left(\sum_{n=0}^\infty \frac n{2^n} + \sum_{n=0}^\infty \frac 1{2^n} \right) = \frac {S_1}2 + \frac {S_0}2 = S_0 = 2 \\ S_2 & = \frac 12 \left(S_2+2S_1+S_0 \right) = 2S_1 + S_0 = 4+2 = 6 \\ S_3 & = \frac 12 \left(S_3+3S_2 + 3S_1 + S_0 \right) = 3 S_2 + 3 S_1 + S_0 = 18+6+2 = 26 \\ S_4 & = \frac 12 \left(S_4+4S_3+6S_2 + 4S_1 + S_0 \right) = 4S_3+6S_2 + 4S_1 + S_0 = 104+36+8+2 = 150 \end{aligned}$

$\implies S = S_4 = 150$ . Since $150= 2 \times 3 \times 5^2$ , $A = (1+1)(1+1)(2+1) = 12$ . Therefore, $S-A = 150-12 = \boxed{138}$ .

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