I have the infinite sum:
$$\sum_{n=1}^{\infty}\frac{1}{2(n+2)}$$
In this sum I observe that all instances of n is added with 2 before used.
Therefor I would think i could do this
$$\sum_{n=1}^{\infty}\frac{1}{2(n+2)} = \sum_{n=3}^{\infty}\frac{1}{2n}$$
Now, I would like it to start at 1 to make it easier to manipulate, so if i move the index down to 1 and then subtract what that gives it should still be equal.
$$\sum_{n=3}^{\infty}\frac{1}{2n} = \left(\sum_{n=1}^{\infty}\frac{1}{2n}\right) - \frac{3}{4}$$
And then I do this:
$$\left(\sum_{n=1}^{\infty}\frac{1}{2n}\right) - \frac{3}{4} = \left(\sum_{n=1}^{\infty}\frac{1}{2}\times\frac{1}{n}\right) - \frac{3}{4} = \left(\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n}\right) - \frac{3}{4}$$
And since $\sum_{n=1}^{\infty}\frac{1}{n}$ diverges, dosen't this mean that the entire sum diverges? Is this a valid proof or are you not allowed to manipulate sums like this if you don't know if it diverges or converges.
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Alice Ryhl
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You don't tell us what level of courses you've taken. Basically everything you've done is fine. How you justify the steps you take depends on what techniques you've learned to manipulate things (but what you've done can be defended no matter what level of formality you're using). – anthonyquas Mar 06 '14 at 19:24
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Informally, convergence or divergence only depends on what happens "out at infinity", so adding or subtracting a constant or any finite number of terms early on doesn't change whether the sum converges.
As long as all the terms in the sum are positive, you can reorder them at will without changing convergence. In a conditionally convergent series you cannot do that.
Your proof is fine. The upper limit of the sum being $\infty$ should be understood as the limit of the sum to $n$ as $n\to \infty$ Until you take the limit, all the manipulations are done on finite series. This is justified.
Ross Millikan
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