# Limit of a series of a sequence [closed]

I have troubles finding the limit of the following series: $\sum_{n=1}^\infty \frac{5+4n-1}{3^{2n+1}}$

So far I figured it'd easier to split the sum into:

$\sum_{n=1}^\infty \frac{5}{3^{2n+1}} \sum_{n=1}^\infty \frac{4n-1}{3^{2n+1}}$

= $\sum_{n=1}^\infty 5 \cdot\frac{1}{3^{2n+1}} +\sum_{n=1}^\infty 4n-1 \cdot \frac{1}{3^{2n+1}}$

And with $\sum_{n=1}^\infty \frac{1}{w^n} = \frac{1}{w-1}$ you get the following terms:

$5\cdot \frac{1}{3^{2n+1}-1} + 4n-1\cdot \frac{1}{3^{2n+1}-1}$

which is bascially a sequence, but im stuck right here.. help is very appreciated!

## closed as off-topic by happy fish, Wjx, Michael Seifert, Silvia, Daniel LichtblauSep 21 '16 at 18:25

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• Is this Question about the Software Mathematica? If so please complement your Question with Code. Else Mathematics or Statistics satisfies your needs better. – user9660 Sep 21 '16 at 11:49

This can be easily investigated with Mathematica.

Define the function

f[n_] := (4 + 4 n)/3^(2 n + 1)


To check if the sum is convergent:

Limit[f[n + 1]/f[n], n -> Infinity]


1/9

and also

Limit[f[n], n -> Infinity]


0

Or just simply

SumConvergence[f[n], n]


True

ListPlot[Table[f[n], {n, Range[10]}], Frame -> True, PlotRange -> All,
PlotStyle -> {PointSize[Medium], Black}, Filling -> Axis,
FrameLabel -> {"n", Rotate["f(n)", 270 Degree]},
PlotLabel -> "Convergence plot"]


To plot the partial sums:

ListPlot[Accumulate@Table[f[n], {n, Range[10]}], Frame -> True,
PlotRange -> All, PlotStyle -> {PointSize[Medium], Black},
FrameLabel -> {"n", "Partial sums"}]


We see that the sum quickly plateaus. Let's finally find the value of the infinite sum:

sum = Sum[f[n], {n, 1, Infinity}]


17/48

N@sum


0.354167

One can do also

NSum[f[n], {n, 1, Infinity}]


0.354167