# Find the partial sum of the series

Case: $$\sum _{n=1}^{\infty}(-1)^{n+1}(2n-1)$$ Question: $$S_n =\ ?$$ My partial solution and investigation:

$$S_n = \sum _{k=1}^n (-1)^{n+1}(2n-1) = 1 -3+5-7+9-11+\cdots+(-1)^{n+1}(2n-1)$$

I see there is a sum of two arithmetic sequences, but I don't know how to take advantage of this fact. After using the formula $S = S_a + S_b = \frac{(a_1+a_n)n}{2} + \frac{(b_1+b_n)n}{2}$ I got a wrong result $-n+(-1)^{n+1}n$.

$$(-1)^{n+1}n$$

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Your approach can be made to work, but you have to count the terms in the two subseries correctly. Suppose first that $n=2m$ is even. Then you have $$1+5+9+\ldots+2(2m-1)-1=1+5+9+\ldots+(4m-3)\;,$$ with $m$ terms, from which you’re subtracting $$3+7+\ldots+(4m-1)\;,$$ also with $m$ terms. The formula for the sum of an arithmetic progression then gives you

$$S_n=\frac{\big(1+(4m-3)\big)m}2-\frac{\big(3+(4m-1)\big)m}2=\frac{-4m}2=-2m=-n=(-1)^{n+1}n\;.$$

Now suppose that $n=2m+1$ is odd; then you have $m+1$ positive terms and $m$ negative terms. The last positive term is $2n-1=2(2m+1)-1=4m+1$, and the last negative term is $4m-1$, so you have

$$S_n=\frac{\big(1+(4m+1)\big)(m+1)}2-\frac{\big(3+(4m-1)\big)m}2=\frac{4m+2}2=2m+1=n=(-1)^{n+1}n\;.$$

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That seems like an awful lot of machinery to use instead of grouping the terms like $(1-3)+(5-7)+(9-11)+\cdots+L = -2 + -2 + -2 +\cdots+L = -2\cdot\bigl\lfloor \frac n2\bigr\rfloor + L$, where $L$ is the leftover term when $n$ is odd. –  MJD Oct 12 '12 at 21:18
@MJD: Of course; were I faced with the problem, I’d most likely use you approach or a proof by induction after calculating a few values. But I was deliberately addressing the difficult that the OP had with this approach, not recommending it; I thought that fairly clear from the first sentence. –  Brian M. Scott Oct 13 '12 at 3:40

This can be easily proved by induction. The formula can be guessed by looking at the partial sums:

$$S_1 = 1, S_2= -2, S_3 = 3, S_4 = -4,..$$

It's easy to see that $S_1=(-1)^{n+1}n|_{n=1} = 1$ , so the formula works for the base case. Now assume the formula holds for arbitrary $n$, i.e. $S_n = (-1)^{n+1}n$. The we have

$$S_{n+1} = S_n + (-1)^{n+2} (2(n+1) -1) \\= (-1)^{n+1} n + (-1)^{n+2} (2n+1) \\= (-1)^{n+1} ( n - 2n -1) \\ = (-1)^{n+1} (-n-1) \\ = (-1)^{n+2} (n+1)$$

which proves the result!

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You don't assume it holds for arbitrary $n$ but that it holds for every $k$ with $1\leq k\leq n$. Assuming it holds for arbitrary $n$ is the same as assuming what you want to prove. –  Pedro Tamaroff Oct 12 '12 at 19:29
Yes, that's what I meant but couldn't express aright –  Ganesh Oct 12 '12 at 19:29

$S_n=1-3+5-7+\cdots$ ($n$ terms ) $=-2-2-\cdots$ ($n/2$ terms) $=-2.(n/2)=-n$, if $n$ is even.

If $n$ is odd then $S_n=-2-2-\cdots(n-1\mbox {terms)+n th term}=(-2)(n-1)/2+(-1)^{n+1}(2n-1)$ i.e. $S_n=-n+1+2n-1=n$

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