Spot a pattern 2

Method 1:

Let $S = 1 + \dfrac{2}{5} + \dfrac{3}{25} + \dfrac{4}{125} + .... = \displaystyle\sum_{n=1}^{\infty} \dfrac{n}{5^{n-1}}.$

Then $S - \left(\dfrac{1}{5} + \dfrac{1}{25} + \dfrac{1}{125} + ....\right) = 1 + \dfrac{1}{5} + \dfrac{2}{25} + \dfrac{3}{125} + .... = 1 + \dfrac{1}{5}\left(1 + \dfrac{2}{5} + \dfrac{3}{25} + ....\right)$

$\Longrightarrow S - \left(\dfrac{1}{5} + \dfrac{1}{25} + \dfrac{1}{125} + ....\right) = 1 + \dfrac{1}{5}S$ , (A).

Now $\dfrac{1}{5} + \dfrac{1}{25} + \dfrac{1}{125} + .... = \dfrac{\dfrac{1}{5}}{1 - \dfrac{1}{5}} = \dfrac{1}{4},$ so equation (A) becomes

$S - \dfrac{1}{4} = 1 + \dfrac{1}{5}S \Longrightarrow \dfrac{4}{5}S = \dfrac{5}{4} \Longrightarrow S = \boxed{\dfrac{25}{16}}.$

Method 2:

In general we have that $\displaystyle\sum_{n=0}^{\infty} x^{n} = \dfrac{1}{1 - x}$ for $|x| \lt 1.$

Differentiating both sides, (the LHS term-by-term), gives us that

$\displaystyle\sum_{n=1}^{\infty} nx^{n-1} = \dfrac{1}{(1 - x)^{2}}.$

But the given series is of the form of the sum on the LHS of this equation with $x = \dfrac{1}{5},$ and thus sums to

$\dfrac{1}{(1 - \frac{1}{5})^{2}} = \boxed{\dfrac{25}{16}}.$

Having forgotten everything about limits - It can't be 1/25 or 16/25 since these are less than 1, and the arithmetic sequence is clearly greater than 1 by the first term. Further, 5/4 - 1 = 1/4. However, the second term 2/5 is greater than 1/4, so the summation must be greater than 5/4. The only option left therefore is 25/16.

This is an arithmetic-geometric series..

Let $\displaystyle S=1+\cfrac { 2 }{ { 5 }^{ 1 } } +\cfrac { 3 }{ { 5 }^{ 2 } } +\cfrac { 4 }{ { 5 }^{ 3 } } +\cfrac { 5 }{ { 5 }^{ 4 } } +\dots =\sum _{ n=1 }^{ \infty }{ \cfrac { n }{ { 5 }^{ n-1 } } } = 5\sum _{ n=1 }^{ \infty }{ \cfrac { n }{ { 5 }^{ n } } }$ .

Therefore we have $\cfrac { S }{ 5 } =\cfrac { 1 }{ { 5 }^{ 1 } } +\cfrac { 2 }{ { 5 }^{ 2 } } +\cfrac { 3 }{ { 5 }^{ 3 } } +\cfrac { 4 }{ { 5 }^{ 4 } } +\dots \\ \cfrac { S }{ 25 } =\cfrac { 1 }{ { 5 }^{ 2 } } +\cfrac { 2 }{ { 5 }^{ 3 } } +\cfrac { 3 }{ { 5 }^{ 4 } } +\cfrac { 4 }{ { 5 }^{ 5 } } +\dots$

Subtracting the two, we get $\cfrac { 4S }{ 25 } =\cfrac { 1 }{ { 5 }^{ 1 } } +\cfrac { 1 }{ { 5 }^{ 2 } } +\cfrac { 1 }{ { 5 }^{ 3 } } +\cfrac { 1 }{ { 5 }^{ 4 } } +\dots =\cfrac { \cfrac { 1 }{ 5 } }{ 1-\cfrac { 1 }{ 5 } } =\cfrac { 1 }{ 4 }$

$\Rightarrow S= \large\color{#20A900}{\boxed{\dfrac{25}{16}}}$