# Calculate Limit of sequence

Calculate: $$\lim_{n\to\infty}\sum\limits_{i=1}^n\frac{2i^2-1}{2^i}=\lim\limits_{n\to\infty}\left(\frac12 + \frac7{2^2} + ... + \frac{2n^2 -1}{2^n}\right)$$

• What is the denominator? What is on the left is not consistent with what is on the right? Commented Jan 5, 2020 at 23:02
• Do you really want to sum on $i$ a function that doesn't involve $i$? As written, the sum on the left side is just $n(2n^2-1)/2^n$, and the limit is zero. [I fear the editor may have messed up, but the original was also messed up, as a limit on $x$ of an expression with no $x$ in it.] Commented Jan 5, 2020 at 23:04
• Write $\frac{2n^2-1}{2^n} = 2 \frac{n^2}{2^n} - \frac{1}{2^n}. \$I couldn't find a similar question on SE, but there is a good answer on a relevant quora question: quora.com/… Commented Jan 5, 2020 at 23:05
• @GerryMyerson I goofed up. My apologies. Commented Jan 5, 2020 at 23:08
• @Adam, math.stackexchange.com/questions/643206/… also math.stackexchange.com/questions/2300889/… and math.stackexchange.com/questions/1072038/… and probably many others. Commented Jan 5, 2020 at 23:16

You can rewrite your sum as $$2\sum_{i=1}^{\infty} i^2 2^{-i} - \sum_{i=1}^{\infty} 2^{-i}$$ Notice that $$\sum_{i=1}^{\infty} x^{-i} = \frac{1}{x-1}$$ from the function's taylor series and the geometric series formula.

Now differentiate this sum to and multiply by $$x$$ to get $$\sum_{i=1}^{\infty}-ix^{-i} = \frac{-x}{(x-1)^2}$$ Repeat the previous step and get $$\sum_{i = 1}^{\infty} i^2 x^{-i} = \frac{x(x+1)}{(x-1)^3}$$ Replace both identities in the first equation using $$x=2$$ and you have $$\frac{2x(x+1)}{(x-1)^3} - \frac{1}{x-1} \Big]_{x=2} = 12 - 1 = 11$$ Hence, $$\sum_{i=1}^{\infty}\frac{2i^2 - 1}{2^i} = 11$$

Lets look at some simpler sums that you should probably know how to work with.

$$S_1 = \sum_\limits{i=1}^n \frac {1}{2^i}\\ S_1 - \frac 12 S_1 = \frac 12 - \frac 14 + \frac 14 - \cdots - \frac {1}{2^{n+1}}\\ S_1 = 1 - \frac {1}{2^{n}}$$

$$S_2 = \sum_\limits{i=1}^n \frac {i}{2^i}\\ S_2 - \frac 12 S_2= \frac 12 - \frac 14 + \frac {2}{4} - \frac 28 + \frac 38 + \cdots\\ S_2 - \frac 12 S_2 =\frac 12+ \sum_\limits{i=2}^n \frac {i-(i-1)}{2^i} - \frac {1}{2^{n+1}}\\ S_2 - \frac 12 S_2 =\sum_\limits{i=1}^n \frac {i}{2^i} - \frac {n}{2^{n+1}}\\ S_2 = 2S_1 - \frac {n}{2^n}$$

Our next one is a little bit trickier.

$$S_3 = \sum_\limits{i=1}^n \frac {(i)(i+1)}{2^i}\\ S_3 - \frac 12 S_3 = \frac 22 - \frac 24 + \frac 64 - \frac 68 + \frac {12}8 - \frac {12}{16} + \cdots\\ S_3 - \frac 12 S_3 = 1 + \sum_\limits{i=2}^n \frac {(i)(i+1) - (i-1)(i)}{2^i} - \frac {n^2-n}{2^{n+1}}\\ S_3 - \frac 12 S_3 = 1 + \sum_\limits{i=2}^n \frac {2i}{2^i} - \frac {n^2+n}{2^{n+1}}\\ S_3 - \frac 12 S_3 = 2S_2 - \frac {n^2+n}{2^{n+1}}\\ S_3 = 4S_2 - \frac {n^2+n}{2^{n}}$$

Now we can find the sum we are looking for as a combination of these pieces.

$$2i^2 - 1 = 2(i)(i+1) - 2i - 1$$

$$\sum_\limits{i=1}^n \frac {2i^2 - 1}{2^i} = \sum_\limits{i=1}^n \frac {2(i)(i+1) - 2i - 3}{2^i} = 2S_3 - 2S_2 - S_1\\ 6S_2 - 2\frac{n^2 + n}{2^n} - S_1\\ 11S_1 - 2\frac{n^2 + n}{2^n} - 6\frac{n}{2^n}\\ 11 - 11\frac{1}{2^n} - 2\frac{n^2 + n}{2^n} - 6\frac{n}{2^n}\\ 11 - \frac{11 +8n + 2n^2}{2^n}$$