Question

alenahelenash · Accepted Answer

To identify the function represented by the given power series:

\sum_{k = 2}^{\infty} \frac{x^{k}}{k (k - 1)}

We can begin by using the formula for the power series representation of a function. Specifically, if

f (x)

has a power series representation centered at

a

, then we have:

f (x) = \sum_{n = 0}^{\infty} c_{n} (x - a)^{n}

where the coefficients

c_{n}

can be calculated using the formula:

c_{n} = \frac{f^{(n)} (a)}{n!}

where

f^{(n)} (a)

denotes the

n

-th derivative of

f

evaluated at

a

.
To apply this formula to our power series, we need to first manipulate it into a form that resembles the power series for a familiar function. In particular, notice that:

\frac{1}{k (k - 1)} = \frac{1}{k - 1} - \frac{1}{k}

Using this fact, we can rewrite our power series as:

\sum_{k = 2}^{\infty} \frac{x^{k}}{k (k - 1)} = \sum_{k = 2}^{\infty} (\frac{x^{k}}{k - 1} - \frac{x^{k}}{k})

Next, we can exchange the order of summation and rewrite the series as:

\sum_{k = 2}^{\infty} \frac{x^{k}}{k - 1} - \sum_{k = 2}^{\infty} \frac{x^{k}}{k} = \sum_{n = 1}^{\infty} \frac{x^{n + 1}}{n} - \sum_{n = 2}^{\infty} \frac{x^{n}}{n}

Notice that the second sum starts at

n = 2

instead of

n = 1

, so we can combine the two sums into one by adjusting the indices:

\sum_{n = 1}^{\infty} \frac{x^{n + 1}}{n} - \sum_{n = 1}^{\infty} \frac{x^{n}}{n} = \sum_{n = 1}^{\infty} \frac{x^{n}}{n} (\frac{x}{n + 1} - 1)

Now, we have a power series that looks like:

\sum_{n = 1}^{\infty} c_{n} x^{n}

where:

c_{n} = \frac{1}{n} (\frac{x}{n + 1} - 1) = \frac{x}{n (n + 1)} - \frac{1}{n}

Thus, we have identified the function represented by the power series as:

f (x) = \sum_{n = 1}^{\infty} (\frac{x}{n (n + 1)} - \frac{1}{n}) x^{n}

which can be simplified further as:

f (x) = \sum_{n = 1}^{\infty} (\frac{x}{n} - \frac{x}{n + 1})

This function represents the difference between the harmonic series and its first term.

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