New Year's Countdown Day 5: Fifth Powers

First, let's examine the limit. Though it looks scary at first, breaking it up into individual fractions will make a pattern emerge:

$\begin{aligned} \lim_{n \rightarrow \infty} \dfrac{5^{1/n} + 32 \cdot 5^{2/n} + 243 \cdot 5^{3/n} + \cdots + n^5 \cdot 5}{n^6} &= \lim_{n \rightarrow \infty} \left ( \dfrac{1^5 5^{1/n}}{n^6} + \dfrac{2^5 5^{2/n}}{n^6} + \dfrac{3^5 5^{3/n}}{n^6} + \cdots + \dfrac{n^5 5^{n/n}}{n^6} \right ) \\ &= \lim_{n \rightarrow \infty} \dfrac{1}{n} \left ( \left ( \dfrac{1}{n} \right)^5 5^{1/n} + \left ( \dfrac{2}{n} \right)^5 5^{2/n} +\left ( \dfrac{3}{n} \right)^5 5^{3/n} + \cdots + \left ( \dfrac{n}{n} \right)^5 5^{n/n} \right ) \\ &= \lim_{n \rightarrow \infty} \sum_{i = 1}^n \left ( \dfrac{i}{n} \right )^5 5^{i/n} \dfrac{1}{n}. \\ \end{aligned}$

We recognize this as the Riemann sum definition of a definite integral . Here, the sum represents the integral of the function $g(x) = x^5 5^x$ over the interval $[0, 1],$ so

$\lim_{n \rightarrow \infty} \dfrac{5^{1/n} + 32 \cdot 5^{2/n} + 243 \cdot 5^{3/n} + \cdots + n^5 \cdot 5}{n^6} = \displaystyle \int_0^1 x^5 5^x dx.$

To find $\displaystyle \int x^5 5^x dx,$ we use a table to organize our integration by parts steps:

$\begin{array}{c|c|c|c} \text{Sign} & u & dv & \text{Term} \\ \hline & & 5^x & \\ + & x^5 & \dfrac{5^x}{\ln 5} & \dfrac{x^55^x}{\ln 5} \\ - & 5x^4 & \dfrac{5^x}{\ln^2 5} & -\dfrac{5x^45^x}{\ln^2 5} \\ + & 20x^3 & \dfrac{5^x}{\ln^3 5} & \dfrac{20x^35^x}{\ln^3 5} \\ - & 60x^2 & \dfrac{5^x}{\ln^4 5} & -\dfrac{60x^25^x}{\ln^4 5} \\ + & 120x & \dfrac{5^x}{\ln^5 5} & \dfrac{120x5^x}{\ln^5 5} \\ - & 120 & \dfrac{5^x}{\ln^6 5} & -\dfrac{120 \cdot 5^x}{\ln^6 5} \end{array}$

Thus,

$\displaystyle \int x^5 5^x dx = \dfrac{x^55^x}{\ln 5} - \dfrac{5x^45^x}{\ln^2 5} + \dfrac{20x^35^x}{\ln^3 5} - \dfrac{60x^25^x}{\ln^4 5} + \dfrac{120x5^x}{\ln^5 5} - \dfrac{120 \cdot 5^x}{\ln^6 5} = \dfrac{5^x((x \ln 5)^5 - 5(x \ln 5)^4 + 20(x \ln 5)^3 - 60(x \ln 5)^2 + 120(x \ln 5) - 120)}{\ln^6 5}.$

Plugging in $x = 0$ yields $-\dfrac{120}{\ln^6 5},$ and plugging in $x = 1$ yields $\dfrac{5(\ln^5 5 - 5\ln^4 5 + 20\ln^3 5 - 60\ln^2 5 + 120\ln 5 - 120)}{\ln^6 5}.$ The difference between the two is equal to

$\displaystyle \int_0^1 x^5 5^x dx = \dfrac{5(\ln^5 5 - 5\ln^4 5 + 20\ln^3 5 - 60\ln^2 5 + 120\ln 5 - 96)}{\ln^6 5},$

which is our final answer for the limit. Comparing the result to the structure of $f,$ we can equate the coefficients and get get $(a_0, a_1, a_2, a_3, a_4) = (96, 120, 60, 20, 5),$ the sum of which comes out to $a_0 + a_1 + a_2 + a_3 + a_4 = \boxed{301}.$