Calculate the sum of a number series

TorioSeduto

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Nov 13, 2021
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Hello everyone, I have the following number series

[math]\sum_{n=1}^{\infty}{\frac{(7n+32) \cdot 3^{n}}{n(n+2) \cdot 4^{n}}}[/math]
I used the ratio criterion and found that the [math] lim_{n\to \infty} {\Biggl| \frac{a_{n+1}}{a_{n}}\Biggr|} \,=\, \frac{3}{4}[/math]
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: [math]\sum_{n=1}^{\infty}{\Biggl( \frac{16}{n} - \frac{9}{n+2}\Biggr) \cdot \Biggr(\frac{3}{4}\Biggr)^{n}}[/math]
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 [imath]\sum_n x_n[/imath] where [imath]x_n = \left(\frac{16}{n}-\frac{9}{n+2}\right)\left(\frac{3}{4}\right)^n[/imath] -- have you looked at [imath]x_n + x_{n+2}[/imath] ?
 
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 [math] 16 \cdot \sum_{n=1}^{\infty}{\frac{1}{n} x^{n}} -9 \cdot \sum_{n=1}^{\infty}{\frac{1}{n+2} x^{n}} [/math]
I take the individual summations and derive them term by term:

[math]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) [/math]
I consider
[math] 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} [/math]consequentially [math] 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}} [/math]
lastly

[math]16 \cdot f_{1}(3/4) - 9\cdot f_{2}(3/4) \,=\, \frac{33}{2} \,=\, 16.5[/math]
 
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This looks cool! And your final results matched the brute force computation!
 
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