True or False question

RHS

We know the following:

$\displaystyle e = \sum_{\color{#3D99F6}k=0}^{\infty} \frac{1}{k!}$

As stated in the question, however, we are dealing with this:

$\displaystyle \sum_{\color{#D61F06}k=1}^{\infty} \frac{1}{k!}$

In order to find the value of this, we take the first term of the series, which is equal to 1. Therefore, the right hand side is equal to $15(e-1) =\color{#20A900} 15e-15$ .

LHS

We shall start of with the equation for $e^x$

$\displaystyle e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!}$

Differentiating wrt x then multiplying by x:

$\displaystyle xe^x = \sum_{k=0}^{\infty} \frac{kx^k}{k!}$

Differentiating wrt x then multiplying by x:

$\displaystyle (x^2+x)e^x = \sum_{k=0}^{\infty} \frac{k^2x^k}{k!}$

Differentiating wrt x then multiplying by x:

$\displaystyle (x^3+3x^2+x)e^x = \sum_{k=0}^{\infty} \frac{k^3x^k}{k!}$

Differentiating wrt x:

$\displaystyle (x^3+6x^2+7x+1)e^x = \sum_{k=0}^{\infty} \frac{k^4x^{k-1}}{k!}$

As the first term is zero, $\displaystyle (x^3+6x^2+7x+1)e^x = \sum_{\color{#3D99F6}k=0}^{\infty} \frac{k^4x^{k-1}}{k!} = \sum_{\color{#D61F06}k=1}^{\infty} \frac{k^4x^{k-1}}{k!}$

Subbing in 1 for $x$ gives us the following:

$\displaystyle \color{#20A900}15e\color{#333333} = \sum_{k=1}^{\infty} \frac{k^4}{k!}$

Final answer

Since it is clear that $\color{#20A900}15e-15\color{#333333}=\color{#20A900}15e$ is false, it is now obvious that the following statement is $\boxed{\color{#D61F06}False}$

$\displaystyle 15\sum_{k=1}^{\infty} \frac{1}{k!} = \sum_{k=1}^{\infty} \frac{k^4}{k!}$

Since $k^4 \; = \; k(k-1)(k-2)(k-3) + 6k(k-1)(k-2) + 7k(k-1) + k$ we have $\sum_{k=1}^\infty \frac{k^4}{k!} \; = \; \sum_{k \ge 4} \frac{1}{(k-4)!} + 6\sum_{k \ge 3} \frac{1}{(k-3)!} + 7\sum_{k \ge 2}\frac{1}{(k-2)!} + \sum_{k \ge 1}\frac{1}{(k-1)!} \; = \; (1+7+6+1)e = 15e$ which is not the same as $15(e-1)$ .