How to find the value this sum converges to?$$\sum_{n=0}^{\infty}n·\frac{3^{n-1}}{4^{3n+1}} $$ I've no clue how to simplify $ \frac{3^{n-1}}{4^{3n+1}} $ so if you could help me with that I could do the rest by myself so any hint would be appreciated.
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It's not a duplicate! That other question talks about $\sum nx^n$,, while here it seems clear that the problem is how to write the sum in that form. – David C. Ullrich Jan 27 '19 at 17:20
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notice that
$$ \frac{ 3^{n-1} }{4^{3n+1} } = \frac{1}{12} \left( \frac{3}{64} \right)^n $$
Now, use the fact that
$$ \sum n x^n = \frac{x}{(1-x)^2} , \; \; \;\ |x|<1$$
as one can easily see that upon differentiating the geometric sum $\sum x^n = \frac{1}{1-x} $ with respect to $x$ and then multiplying by $x$ and since the radius of convergence remains the same upon differentition and the result follows
James
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