Sums of Infinite Series - Pre-Calculus

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Question

Find the value for $\sum_{i=0}^{\infty} (\frac{1}{2})^i$

Answer

To best understand, let's write out the series. So

$\sum_{i=0}^{\infty} (\frac{1}{2})^i = 1+1/2+(1/2)^2 + (1/2)^3+...+(1/2)^\infty$

We can see this is an infinite geometric series with each successive term being multiplied by $\frac{1}{2}$ .

A definition you may wish to remember is

$1+r+r^2+r^3+... +r^\infty= 1/(1-r)$ where stands for the common ratio between the numbers, which in this case is $\frac{1}{2}$ or . So we get

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Question

Evaluate:

$1 - \frac{1}{6} + \frac{1}{36} - \frac{1}{216} +...$

Answer

This is a geometric series whose first term is $a_{0} = 1$ and whose common ratio is $r = - \frac{1}{6}$ . The sum of this series is:

$\frac{a_{0}}{1-r} = \frac{1}{1- \left (- \frac{1}{6} \right )} = \frac{1}{\frac{6}{6}+ \frac{1}{6}}= \frac{1}{\frac{7}{6}} =\frac{6}{7}$

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Question

Evaluate:

$1 + \frac{1}{7} + \frac{1}{49} + \frac{1}{343} +...$

Answer

This is a geometric series whose first term is $a_{0} = 1$ and whose common ratio is $r = \frac{1}{7}$ . The sum of this series is:

$\frac{a_{0}}{1-r} = \frac{1}{1- \frac{1}{7} } = \frac{1}{\frac{7}{7}- \frac{1}{7}}= \frac{1}{\frac{6}{7}} =\frac{7}{6}$

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Question

What is the sum of the following infinite series?

$S=2-\frac{1}{3}+1-\frac{1}{9} +\frac{1}{2}-\frac{1}{27}+\frac{1}{4}-\frac{1}{81}+\cdots$

Answer

This series is not alternating - it is the mixture of two geometric series.

The first series has the positive terms.

$2+1+\frac{1}{2}+ \frac{1}{4} + \cdots = 2\sum\limits_{i=0}^\infty (\frac{1}{2})^i = 2(\frac{1}{1-\tfrac{1}{2}})=4$

The second series has the negative terms.

$-\frac{1}{3}- \frac{1}{9} -\frac{1}{27}- \cdots = -\frac{1}{3}\sum\limits_{i=0}^\infty (\frac{1}{3})^i = -\frac{1}{3}(\frac{1}{1-\tfrac{1}{3}})=-0.5$

The sum of these values is 3.5.

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Question

What is the sum of the alternating series below?

$10 - 5 + \frac{5}{2} - \frac{5}{4}+\frac{5}{8}-\frac{5}{16}+\cdots$

Answer

The alternating series follows a geometric pattern.

$10 - 5 + \frac{5}{2} - \frac{5}{4} + \cdots = 10\sum\limits_{i=0}^\infty(-\frac{1}{2})^i$

We can evaluate the geometric series from the formula.

$10\sum\limits_{i=0}^\infty(-\frac{1}{2})^i = 10\frac{1}{1+\tfrac{1}{2}} = 10\frac{2}{3} =\frac{20}{3}$

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Question

Find the sum of the following infinite series:

$9 + 3 + 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{81} + \frac{1}{243} + .....$

Answer

Notice that this is an infinite geometric series, with ratio of terms = 1/3. Hence it can be rewritten as:

$S = \sum_{n=0}^{\infty } 9\cdot \bigg(\frac{1}{3}\bigg)^{n}$

Since the ratio, 1/3, has absolute value less than 1, we can find the sum using this formula:

$S = \frac{a_{0}}{1 - r}$

Where is the first term of the sequence. In this case $a_0 = 9, r = \frac{1}{3}$ , and thus:

$\frac{9}{1-\frac{1}{3}} = \frac{9}{\frac{2}{3}} = \frac{9}{1}\cdot \frac{3}{2}=\frac{27}{2} = 13.5$

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Question

In the infinite series $a_{1}, a_{2}...a_{n},$ each term $a_{n}=(-2)^{n}$ such that the first two terms are and . What is the sum of the first eight terms in the series?

Answer

Once you're identified the pattern in the series, you might see a quick way to perform the summation. Since the base of the exponent for each term is negative, the result will be positive if is even, and negative if it is odd. And the series will just list the first 8 powers of 2, with that positive/negative rule attached. So you have:

-2, 4, -8, 16, -32, 64, -128, 256

Note that each "pair" of adjacent numbers has one negative and one positive. for the first pair, -2 + 4 = 2. For the second, -8 + 16 = 8. For the third, -32 + 64 = 32. And so for the fourth, -128 + 256 = 128. You can then quickly sum the values to see that the answer is 170.

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