Calculate the sum of a number series

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TorioSeduto

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Hello everyone, I have the following number series

∑n=1∞(7n+32)⋅3nn(n+2)⋅4n\sum_{n=1}^{\infty}{\frac{(7n+32) \cdot 3^{n}}{n(n+2) \cdot 4^{n}}}
I used the ratio criterion and found that the limn→∞∣an+1an∣ = 34 lim_{n\to \infty} {\Biggl| \frac{a_{n+1}}{a_{n}}\Biggr|} \,=\, \frac{3}{4}
Knowing that the series converges, at this point, to calculate the sum of the series, I thought of using the partial fraction decomposition and I rewrote the series as: ∑n=1∞(16n−9n+2)⋅(34)n\sum_{n=1}^{\infty}{\Biggl( \frac{16}{n} - \frac{9}{n+2}\Biggr) \cdot \Biggr(\frac{3}{4}\Biggr)^{n}}
At this point how can I calculate the sum of the series?
 
It seems to me that you are the right track. So you are computing ∑nxn\sum_n x_n where xn=(16n−9n+2)(34)nx_n = \left(\frac{16}{n}-\frac{9}{n+2}\right)\left(\frac{3}{4}\right)^n -- have you looked at xn+xn+2x_n + x_{n+2} ?
 
Have you looked at the 1st few terms? This might help you see what is going on!
 
Update, maybe I understand how to proceed in carrying out the exercise.

I consider the series as a power series so I rewrite it as 16⋅∑n=1∞1nxn−9⋅∑n=1∞1n+2xn 16 \cdot \sum_{n=1}^{\infty}{\frac{1}{n} x^{n}} -9 \cdot \sum_{n=1}^{\infty}{\frac{1}{n+2} x^{n}}
I take the individual summations and derive them term by term:

f1(x)=∑n=1∞1nxn ⟶ f1′(x)=∑n=1∞xn−1 = 11−x ⟶ f1(x) = −log(1−x)f_{1}(x) = \sum_{n=1}^{\infty}{\frac{1}{n} x^{n}} \, \longrightarrow \, f'_{1}(x) = \sum_{n=1}^{\infty}{x^{n-1}} \,=\, \frac{1}{1-x} \, \longrightarrow \, f_{1}(x) \,=\, -log(1-x)
I consider
g2(x) = x2⋅f2(x) = x2⋅∑n=1∞1n+2xn = ∑n=1∞1n+2xn+2⟶ g2′(x) = ∑n=1∞xn+1 = x2⋅∑n=0∞xn = x21−x g_{2}(x) \,=\, x^{2} \cdot f_{2}(x) \,=\, x^{2} \cdot \sum_{n=1}^{\infty}{\frac{1}{n+2}}x^{n} \,=\, \sum_{n=1}^{\infty}{\frac{1}{n+2}}x^{n+2} \longrightarrow \, g'_{2}(x) \,=\, \sum_{n=1}^{\infty}{x^{n+1}} \,=\, x^{2} \cdot \sum_{n=0}^{\infty}{x^{n}} \,=\, \frac{x^{2}}{1-x} consequentially g2(x) = −x−x22−log(1−x) ⟶ f2(x) = −1x−12−log(1−x)x2 g_{2}(x) \,=\, -x -\frac{x^{2}}{2} - log(1-x) \, \longrightarrow \, f_{2}(x) \,=\, -\frac{1}{x} - \frac{1}{2} - \frac{log(1-x)} {x^{2}}
lastly

16⋅f1(3/4)−9⋅f2(3/4) = 332 = 16.516 \cdot f_{1}(3/4) - 9\cdot f_{2}(3/4) \,=\, \frac{33}{2} \,=\, 16.5
 
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This looks cool! And your final results matched the brute force computation!
 
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