Find the limit as x approaches infinity of the sum from y = 1 to x of y divided by 2 to the power y.

Step-by-step solution

1. 1

   Recognise this as an infinite series, not a limit of a function. The partial sums are increasing and the terms $y/2^y$ shrink geometrically, so the limit is exactly the value of the convergent series

   $S=\sum_{y=1}^{\infty}\frac{y}{2^y}$

   Convergence is guaranteed by the ratio test: the ratio of consecutive terms tends to $\tfrac12<1$.

2. 2

   Write out the sum and halve it.

   $S=\frac12+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+\cdots$

   $\frac12 S=\frac14+\frac{2}{8}+\frac{3}{16}+\frac{4}{32}+\cdots$

   Halving shifts every term one place right, which is what makes the next step work.

3. 3

   Subtract the two rows. Aligning by denominator, each coefficient drops by exactly one:

   $S-\frac12 S=\frac12+\frac{2-1}{4}+\frac{3-2}{8}+\frac{4-3}{16}+\cdots=\frac12+\frac14+\frac18+\cdots$

   The messy arithmetico-geometric series has become a plain geometric one.

4. 4

   Sum the geometric series and solve. With first term $\tfrac12$ and ratio $\tfrac12$:

   $\frac12+\frac14+\frac18+\cdots=\frac{1/2}{1-1/2}=1$

   so $\tfrac12 S=1$ and therefore $S=2$.

5. 5

   Cross-check with the closed form and numerically. The standard identity $\sum_{y\ge1}yr^y=\dfrac{r}{(1-r)^2}$ at $r=\tfrac12$ gives $\dfrac{1/2}{1/4}=2$ ✓. Adding the first $199$ terms directly gives $2.0000000000$ to ten decimal places ✓.