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**I Don't understand what's The Relation or difference between Infinite series and limits **

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  • $\begingroup$ Welcome to Mathematics Stack Exchange. Infinite series are evaluated as the limit of the infinite sequence of partial sums $\endgroup$ – J. W. Tanner May 15 at 20:39
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    $\begingroup$ Could you explain more $\endgroup$ – Omar Ahmed Ph May 15 at 20:40
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    $\begingroup$ $\sum_{k=0}^\infty a_k=\lim_{n\to\infty} \sum_{k=0}^n a_k$ $\endgroup$ – J. W. Tanner May 15 at 20:42
  • $\begingroup$ What is the relation between Limits and Summation notation $\endgroup$ – Omar Ahmed Ph May 15 at 20:43
  • $\begingroup$ infinite sum is evaluated as limit of finite (partial) sums $\endgroup$ – J. W. Tanner May 15 at 20:46
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An infinite sum is a limit.


The notation $$\sum_{i=1}^\infty a_i$$ (for example) is a shorthand for a limit expression. Specifically, it means $$\lim_{n\rightarrow\infty}S_n,$$ where $S_n$ is the $n$th partial sum $$\sum_{i=1}^na_i.$$

More concisely, the definition of infinite summation is $$\sum_{i=1}^\infty a_i=\lim_{n\rightarrow\infty}\sum_{i=1}^n a_i.$$

These finite sums $\sum_{i=1}^na_i$ have nothing to do with limits, they're just concrete objects.


For example, let's look at $$\sum_{i=1}^\infty 2^{-i}.$$ Here $a_i=2^{-i}$. Intuitively our sum is $${1\over 2}+{1\over 4}+{1\over 8}+{1\over 16}+...,$$ which "clearly" equals $1$. Let's make sure this is right.

We compute the partial sums: $$S_n=\sum_{i=1}^n 2^{-i}={1\over 2}+{1\over 4}+...+{1\over 2^n}=1-{1\over 2^n}.$$

(You should make sure you're comfortable with this kind of calculation before tackling infinite series.)

Taking the limit, we have $$\lim_{n\rightarrow\infty} S_n=\lim_{n\rightarrow\infty}(1-{1\over 2^n})=1.$$

So the answer is $$\color{red}{\sum_{i=1}^\infty 2^{-i}=1}$$ which is exactly what we hoped we'd get.

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