Sum notation confusion

Question

Consider the following sum:

$$\sum_{i=0}^n e^{i/n}$$

I don't understand this notation. Apparently the closed form is $$\dfrac{e^{(n+1)/n }-1}{e^{1/n} -1}$$

But it says $i=0$. I really don't understand how $i$ and $n$ can be in the sum at the same time.

Answer 1

The notation seems to stand for,

$$\sum_{i=0}^n e^{i/n} = e^{0} + e^{\frac{1}{n}} + e^{\frac{2}{n}} +... + e^{\frac{n}{n}} \tag{1}$$

To find the closed form multiply $(1)$ by $e^{\frac{1}{n}} $ to get $$ e^{\frac{1}{n}} \sum_{i=0}^n e^{i/n} = e^{\frac{1}{n}} + e^{\frac{2}{n}} + e^{\frac{3}{n}} +... + e^{\frac{n + 1}{n}} \tag{2}$$

$(2) - (1)$ yields the result.

$*** \; $Note that $ e^{0} = 1$.

And verify that $e^{\frac{1}{n}} - 1 \gt 0 $ by the series expansion of $e^x$.

Answer 2

The notation $\sum_{x=l}^h f(x)$ reads "sum $f(x)$ for $x$ from $l$ to $h$". We tacitly assume that $l,h$ are integers and $l \leq h$ (most of the time at least). In that case, $$ \sum_{x=l}^h f(x) = x(l) + x(l+1) + \cdots + x(h). $$ For example, $$ \sum_{t=-1}^2 t^2 = (-1)^2 + 0^2 + 1^2 + 2^2 = 6. $$

Answer 3

$i$ is just a ``dummy variable of summation.'' In general, $$ \sum_{i=0}^{n}f_{i}=f_{0}+f_{1}+\ldots+f_{n}. $$ In your case, $$ f_{i}\equiv e^{i/n}. $$ If we substitute, we get $$ \sum_{i=0}^{n}e^{i/n}=e^{0/n}+e^{1/n}+\ldots+e^{n/n}. $$

There is a general result about geometric series, which are summations that look like: $$ \sum_{i=0}^{n}x^{i} $$ This result is that $$ \sum_{i=0}^{n}x^{i}=\frac{x^{n+1}-1}{x-1}. $$ In your case, $x=e^{1/n}$. Substituting into the above $$ \frac{e^{\left(n+1\right)/n}-1}{e^{1/n}-1} $$

Answer 4

But it says $i=0$ . I really don't understand how $i$ and $n$ can be in the sum at the same time.

I guess you are familiar with summations such as: $\displaystyle\sum\limits_{n=a}^b x^n = (x^a+x^{a+1}+\cdots +x^b) = \dfrac{x^{b+1}-x^a}{x-1}$

Here $n$ is the series index. As such its scope is the series; it only occurs within the terms of the series, never in the evaluation of the sum. You can thus replace it with any symbol, as long as it is not already in use. You don't have to use $n$.

Hence this is exactly the same series : $\displaystyle\sum\limits_{k=a}^b x^k = (x^a+x^{a+1}+\cdots +x^b) = \dfrac{x^{b+1}-x^a}{x-1}$

Here we use $k$ as the series index. $k$ is iterated from $a$ to $b$ exactly as $n$ was in the first expression, and the term $x^k$ takes the values $x^a, x^{a+1}, \ldots, x^b$ exactly as $x^n$ did in the first expression.

So if you have the series summed from $0$ to $n$, you cannot use $n$ as the series index when it's used as the upper bound. So you must use another symbol; such as $i$.

$$\sum\limits_{i=0}^n x^i = \dfrac{x^{n+1}-1}{x-1}$$

You can also make use of $n$ inside the term. for instance, replacing $x$ with $e^{1/n}$

$$\sum\limits_{i=0}^n e^{i/n} = \dfrac{e^{(n+1)/n}-1}{e^{1/n}-1}$$

So, in summary: You can use any symbol as the series index as long as it is not claimed elsewhere in the expression, such as the term or the bounds. The series index, whatever it is, is iterated from the lower bound to the upper bound. Only the series index is so iterated; if $n$ occurs in the expression and is not the series index, it is not iterated.