Proof of Convergence/Divergence Unaffected by Series Starting Point

AI Thread Summary
The discussion centers on the convergence of infinite series, asserting that if a series converges from an integer T to infinity, it also converges from 1 to infinity, and vice versa. This principle is derived from the definition of convergent series, although some caution is advised regarding specific examples. A counterexample is provided with the series from x=1, which does not converge due to an undefined term at x=7. The conversation highlights the importance of careful wording and understanding the conditions under which convergence applies. Overall, the starting point of a series does not affect its convergence or divergence in general cases, but exceptions exist.
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if we know that an infinite series is convergent from an integer T, to infinity, then the series is convergent from 1 to infinity. conversely, if a series is convergent from 1 to infinity then it is convergent from T to infinity (i.e. starting point of the series does not affect convergence/divergence) This seems obvious but can anyone help me prove it please.
 
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This is obvious from the definition of a convergent series. So check the definition again.
 
I pulled my lecturer up on this recently, consider the series:

\sum_{x=10}^{\infty} \frac{1}{(x-7)^2}

It converges and if you are interested to:

\frac{1}{12} \left(2 \pi^2 - 15\right)

However:

\sum_{x=1}^{\infty} \frac{1}{(x-7)^2}

Clearly does not converge, so be careful how you word it. Anyway, it's not too difficult to prove, just think of it like:

a_1 + a_2 + \ldots + a_{t-1} + \sum_{n=t}^{\infty} a_n
 
Is the series adamg asking about a power series? If so, the answer to his question is yes. However in the more general case as Zurtex showed, it is not true.
 
I'd rather say his example

\sum_{x=1}^{\infty} \frac{1}{(x-7)^2}

is not a series, since the 7'th term is not defined.
 
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