Question

How do you find the sum of infinite series ${{\sum{\left( \dfrac{1}{10} \right)}}^{k}}$ from $k=1\text{ to }\infty$ ?

Answer 1

From the question we need to find the sum of the infinite geometric series $\sum{{{\left( \dfrac{1}{10} \right)}^{k}}}$ from $k=1\text{ to }\infty$ . We know that the formula for finding the sum of the infinite geometric series is given as $\dfrac{a}{1-r}$ where $a$ is the first term and $r$ is the common ratio.

Complete step by step solution:
Now considering from the question we have been asked to find the sum of infinite geometric series $\sum{{{\left( \dfrac{1}{10} \right)}^{k}}}$ from $k=1\text{ to }\infty$ .
From the basics of series concepts we know that the formula for finding the sum of the infinite geometric series is given as $\dfrac{a}{1-r}$ where $a$ is the first term and $r$ is the common ratio.
Here in the given series the first term is $\dfrac{1}{10}$ and the common ratio is $\dfrac{1}{10}$ .
Hence the sum of the infinite geometric series is given as $\Rightarrow \dfrac{\left( \dfrac{1}{10} \right)}{\left( 1-\dfrac{1}{10} \right)}=\dfrac{1}{9}$ .
Therefore we can conclude that the sum of the given infinite geometric series ${{\sum{\left( \dfrac{1}{10} \right)}}^{k}}$ from $k=1\text{ to }\infty$ is given as $\dfrac{1}{9}$ .

Note: In the process of solving questions of this type we should be sure with our concepts that we are going to apply if we have a clear concept then we can answer them in a short span of time accurately. Similarly we have formula for finding the sum of $n$ terms in geometric series given as $\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}$ and for arithmetic series the formula is given as $\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$ where $d$ is the common difference and $a$ is the first term. The ${{n}^{th}}$ term of geometric series is given as $a{{r}^{n-1}}$ and for arithmetic series $a+\left( n-1 \right)d$ .