Limit of series with exponent

Question

I want to calculate the limit of the following series:

$$ \sum^{\infty}_{k=0} \frac{(-3)^k +5}{4^k}$$

My first step would be to split the term into these parts:

$ \sum^{\infty}_{k=0} \frac{(-3)^k}{4^k}$ $ \sum^{\infty}_{k=0} \frac{5}{4^k}$

If both of them have a limit I can just add them together, right ?

I have looked through my notes on limits and convergence but I dont know how to get rid of the exponent so I can determine the limit.

I have used various online calculators but I could not understand their result.

Answer 1

Do you know about geometric series?

In this problem, you have $$ \sum_{k=0}^{\infty} \frac{(-3)^k}{4} = \sum_{k=0}^{\infty} \left( \frac{-3}{4} \right)^k = \frac{1}{1-\left(\frac{-3}{4}\right)}, $$ and $$ \sum_{k=0}^{\infty} \frac{5}{4^k} = 5 \sum_{k=0}^{\infty} \left( \frac{1}{4} \right)^k = 5 \cdot \frac{1}{1-\frac{1}{4}}$$ Simplify and add to get what you want.

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A quick, not entirely rigorous justification: suppose that $|r| < 1$, and that we want to sum $$ \sum_{k=0}^{\infty} r^k. $$ Since $|r|<1$, we know that this series converges by the ratio test. So, suppose that the limit is $S$; that is $$ S := \sum_{k=0}^{\infty} r^k = 1 + r + r^2 + r^3 + \cdots. $$ Multiplying by $r$, we get $$ rS = r + r^2 + r^3 + r^4 + \cdots .$$ Subtracting, we have \begin{align} &S - rS = (1+r+r^2+\cdots) - (r+r^2+r^3+\cdots) \\ &\qquad\implies (1-r)S = 1 + (r-r) + (r^2-r^2) + \cdots = 1 \\ &\qquad\implies S = \frac{1}{1-r}, \end{align} which is the result used above. Note that the second line is justified, as series involved all converge absolutely, and so we may rearrange the terms without being too careful.

Answer 2

Yes, that's a good first step (assuming you're adding the two sums).

Now rewrite:

$$\sum_{k=0}^{\infty} \left(\frac{-3}{4}\right)^k + 5\sum_{k=0}^{\infty}\frac{1}{4^k}$$

Both of these are the sums of geometric series, so plugging in to the formula we find:

$$\frac{1}{1 - \left(\frac{-3}{4}\right)} + \frac{5}{1 - \frac{1}{4}}$$ $$ = \frac{4}{7} + \frac{20}{3} = \frac{132}{21}$$

Note that both ratios fall within the radius of convergence, so these sums do indeed converge and we can use the geometric series formula.

Answer 3

Hint: $\sum_na^n, |a|<1$ is ${1\over{1-a}}$, take $a=({{-3}\over 4})$ and $b={1\over 4}$ multiply the sum of the last serie by $5$ and add the both results.

Answer 4

You'll need to use the fact that, for $|r| < 1$,

$$ \sum_{j \geq 0} r^j = \frac{1}{1-r}. $$

The first sum can be written as

\begin{align*} \sum_{k \geq 0} \frac{(-3)^k}{4^k} &= \sum_{k \geq 0} \left( -\frac{3}{4} \right) ^k \\ &= \frac{1}{1+\frac{3}{4}} \\ &= \frac{4}{7} \end{align*}

The second is similar, where, after pulling the constant out,

$$ 5\sum_{k \geq 0} \left( \frac{1}{4} \right) ^k = 5 \cdot \frac{4}{3} $$

(check this).

Both limits exist, and so the limit of the sum (of sums) is the sum of limits.

Answer 5

The sum of a geometric series:

$\sum_{i=0}^n a^i\\ \frac {(1-a)}{(1-a)} \sum_{i=0}^n a^i$

Multiply the numerator, term by term into the series, and you see the series telescopes.

$1-a+a-a^2+a^2 - a^3\cdots + a^{n} - a^{n+1}$

$\sum_{i=0}^n a^i = \frac {1-a^{n+1}}{1-a}$

If $|a|<1$ we can take the limit as $n$ approaches infinity.

$\lim_\limits{n\to\infty} \sum_{i=0}^n a^i = \frac {1}{1-a}$

You will sometimes see this proof:

Suppose, $S = \sum_{i=0}^{\infty} a^i$ coverges.

$S - aS = \sum_{i=0}^{\infty} a^i - \sum_{i=1}^{\infty} a^i = 1\\ S = \frac {1}{1-a}$

And then prove that the series converges when $|a| < 1$