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I am struggling a bit with power series at the moment, and I don't quite understand what this question is asking me to do? Am I meant to form a power series from these, or simply evaluate that series? Any explanation/working is appreciated.

Using power series representation, calculate

$$\sum_{n=1}^\infty \frac{n2^n}{3^n}.$$

Daniel Fischer
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user109122
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4 Answers4

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Recall that, in general, $$1 + x + x^2 + \cdots = \frac{1}{1 - x}, \quad |x| < 1.$$

Moreover, power series can be differentiated term by term. So, differentiating both sides of the equation above we get $$1 + 2x + 3x^2 + \cdots = \frac{1}{(1 - x)^2}, \quad |x| < 1.$$

Now, multiplying both sides by $x$ leads to $$x + 2x^2 + 3x^3 + \cdots = \sum_{n = 1}^\infty nx^n = \frac{x}{(1 - x)^2}, \quad |x| < 1.$$

However, in this case $x = 2/3 < 1$, so simply substitute $x = 2/3$ in formula above.

glebovg
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Hint: Define $f$ by

$$f(x) = \sum_{n = 1}^{\infty} x^n$$

Using the formula for a geometric series, this is the same as

$$f(x) = \frac{x}{1 - x}$$

Now compute $f'$ two different ways.

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I believe this Youtube video does a great job explaining the issue. Basically you start with

$$ \sum_{x = 0}^{\infty} x ^ n = \frac{1}{1 - x}.$$

And then you do all the mathematical operations such as $ \frac{d}{dx} $ on both sides until you get the form you want. For example, the first derivative will give you

$$ \sum_{x = 1}^{\infty} n x ^ {n-1} = \frac{1}{(1 - x) ^ 2}.$$

A popular second step you can do from there is multiply both sides by $x$, which gives you

$$ \sum_{x = 1}^{\infty} n x ^ {n} = \frac{x}{(1 - x) ^ 2}.$$

Waldir Leoncio
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First observe that your series is the special case of

$$\sum_{n=1}^\infty n z^n$$

with $z = \frac{2}{3}$, which has radius of convergence $R=1$.

By using the Cauchy product on $\sum_{n=0}^\infty z^n = \frac{1}{1-z}$ we get

$$\big( \frac{1}{1-z} \big)^2 = \Big( \sum_{n=0}^\infty z^n \Big)^2 = \sum_{n=0}^\infty \Big( \sum_{k=0}^n \Big) z^n = \sum_{n=0}^\infty (n+1) z^n = \sum_{n=1}^\infty n z^{n-1}$$

and after multiplying by $z$

$$\sum_{n=1}^\infty n z^n = \frac{z}{(1-z)^2}$$

For $z = \frac{2}{3}$ we get $\frac{\frac{2}{3}}{(1-\frac{2}{3})^2} = 2 \cdot 3 = 6$.

mdcq
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