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When we have two convergent series that looks like:

$$\sum_{n=m}^\infty a_n=\sum_{n=m}^\infty b_n$$

Can we than conclude that?:

$$a_n=b_n$$

  • 0
    Does the equality hold for all $m\geq 1$, or just a particular $m$? – 2017-01-05
  • 0
    All values of $m$ – 2017-01-05

2 Answers 2

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If the equality holds for all $m$, then yes. This is because

$$a_1 = \sum_{n=1}^\infty a_n - \sum_{n=2}^\infty a_n = \sum_{n=1}^\infty b_n - \sum_{n=2}^\infty b_n=b_1$$

and

$$a_2 = \sum_{n=2}^\infty a_n - \sum_{n=3}^\infty a_n = \sum_{n=1}^\infty b_n - \sum_{n=2}^\infty b_n=b_2$$

and

$$a_3 = \sum_{n=2}^\infty a_n - \sum_{n=3}^\infty a_n = \sum_{n=1}^\infty b_n - \sum_{n=2}^\infty b_n=b_3$$

and so on.

In general,

$$a_k = \sum_{n=k}^\infty a_n - \sum_{n=k+1}^\infty a_n=\sum_{n=k}^\infty b_n - \sum_{n=k+1}^\infty b_n = b_k$$

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Since convergent series can be subtracted term-by-term, you can say

$\hspace{.15 in}\displaystyle a_m=\sum_{n=m}^{\infty}a_n-\sum_{n=m+1}^{\infty}a_n=\sum_{n=m}^{\infty}b_n-\sum_{n=m+1}^{\infty}b_n=b_m\;$ for all $m$