# What is the value of 1/n sum_{k=1}^n e^{k/n} ?

Dec 19, 2015

$e - 1$ (after taking the limit as $n \to \infty$)

#### Explanation:

${\sum}_{k = 1}^{n} {e}^{\frac{k}{n}} = {\sum}_{k = 0}^{n - 1} {e}^{\frac{1}{n}} {e}^{\frac{k}{n}}$

$= {\sum}_{k = 0}^{n - 1} {e}^{\frac{1}{n}} {\left({e}^{\frac{1}{n}}\right)}^{k}$

By the geometric sum formula
${\sum}_{k = 0}^{n - 1} a {r}^{k} = a \frac{1 - {r}^{n}}{1 - r}$
we have

${\sum}_{k = 1}^{n} {e}^{\frac{k}{n}} = {e}^{\frac{1}{n}} \frac{1 - {\left({e}^{\frac{1}{n}}\right)}^{n}}{1 - {e}^{\frac{1}{n}}}$

$= {e}^{\frac{1}{n}} \frac{1 - e}{1 - {e}^{\frac{1}{n}}}$

And so

$\frac{1}{n} {\sum}_{k = 1}^{n} {e}^{\frac{k}{n}} = {e}^{\frac{1}{n}} \frac{1 - e}{n \left(1 - {e}^{\frac{1}{n}}\right)}$

To evaluate the above as $n \to \infty$, note that

${e}^{\frac{1}{n}} \frac{1 - e}{n \left(1 - {e}^{\frac{1}{n}}\right)} = {e}^{\frac{1}{n}} \left(1 - e\right) \cdot \frac{1}{n \left(1 - {e}^{\frac{1}{n}}\right)}$

Thus, if
${\lim}_{n \to \infty} {e}^{\frac{1}{n}} \left(1 - e\right)$
and
${\lim}_{n \to \infty} \frac{1}{n \left(1 - {e}^{\frac{1}{n}}\right)}$
both converge, then

${\lim}_{n \to \infty} {e}^{\frac{1}{n}} \frac{1 - e}{n \left(1 - {e}^{\frac{1}{n}}\right)} = {\lim}_{n \to \infty} {e}^{\frac{1}{n}} \left(1 - e\right) \cdot {\lim}_{n \to \infty} \frac{1}{n \left(1 - {e}^{\frac{1}{n}}\right)}$

(*)

By direct substitution,

${\lim}_{n \to \infty} {e}^{\frac{1}{n}} \left(1 - e\right) = {e}^{\frac{1}{\infty}} \left(1 - e\right)$

$= {e}^{0} \left(1 - e\right)$

$= 1 - e$

Unfortunately, we cannot use direct substitution on
${\lim}_{n \to \infty} \frac{1}{n \left(1 - {e}^{\frac{1}{n}}\right)}$
as this gives us $0 \cdot \infty$ in the denominator.
Instead, we will modify the expression so we can use l'Hopital's rule.

${\lim}_{n \to \infty} \frac{1}{n \left(1 - {e}^{\frac{1}{n}}\right)} = {\lim}_{n \to \infty} \frac{\frac{1}{n}}{1 - {e}^{\frac{1}{n}}}$ which gives us the $\frac{0}{0}$ form needed to apply l'Hopital's rule. Doing so, we have

${\lim}_{n \to \infty} \frac{\frac{1}{n}}{1 - {e}^{\frac{1}{n}}} = {\lim}_{n \to \infty} \frac{\frac{d}{\mathrm{dx}} \left(\frac{1}{n}\right)}{\frac{d}{\mathrm{dx}} \left(1 - {e}^{\frac{1}{n}}\right)}$

$= {\lim}_{n \to \infty} \frac{- \frac{1}{n} ^ 2}{{e}^{\frac{1}{n}} / {n}^{2}}$

$= {\lim}_{n \to \infty} - \frac{1}{e} ^ \left(\frac{1}{n}\right)$

We can now evaluate this by direct substitution.

${\lim}_{n \to \infty} - \frac{1}{e} ^ \left(\frac{1}{n}\right) = - \frac{1}{e} ^ \left(\frac{1}{\infty}\right)$

$= - \frac{1}{e} ^ 0$

$= - 1$

Meaning we have ${\lim}_{n \to \infty} \frac{1}{n \left(1 - {e}^{\frac{1}{n}}\right)} = - 1$

So by our initial statement (*)

${\lim}_{n \to \infty} {e}^{\frac{1}{n}} \frac{1 - e}{n \left(1 - {e}^{\frac{1}{n}}\right)} = \left(1 - e\right) \cdot \left(- 1\right) = e - 1$