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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 $-n+(-1)^{n+1}n$.
The correct answer:

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

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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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    @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.2012-10-13
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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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    Yes, that's what I meant but couldn't express aright2012-10-12
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$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 $