1. If $\displaystyle \left|\sum_{k=0}^\infty a_k\right|\leq M$ Where $M$ is a number, how can you argue that the series is convergent?
  2. If $\displaystyle \sum_{k=0}^\infty a_k=M$ (so your series is convergent), Can you say that $\displaystyle \sum_{k=n+1}^\infty a_k$ also converges?

2 Answers 2


1) The question doesn't make sense as stated, since in stating the condition you're already assuming that the series converges (else the sum to infinity would be undefined). The question would make sense if you consider the partial sums instead, i.e.

$$\left|\sum_{k=0}^na_k\right|\le M$$

for all $n$. In this case, you have a bounded sequence, and you can conclude that it converges if it is monotonic, which is to say that all the terms of the series have the same sign.

2) Yes you can; convergence depends only on what happens "at infinity", and you can always change or drop a finite number of terms without changing convergence. In fact, you can calculate the limit as

$$\sum_{k=n+1}^\infty a_k = M - \sum_{k=0}^n a_k\;.$$

  • $\begingroup$ +1 for the concise answer, I guess it appeared while I was typing mine. $\endgroup$
    – Gerben
    Mar 20, 2011 at 11:37
  • 1
    $\begingroup$ One might add Re 2: ...hence the sequence of the rests converges to zero. $\endgroup$
    – Did
    Mar 20, 2011 at 14:07
  1. By writing your inequality this way, you're already supposing that the series converges; otherwise, $\sum_{k=0}^{\infty} a_k$ has no meaning.
  2. Yes. Morally, $\sum_{k=n+1}^{\infty} a_k = M - \sum_{k=0}^n a_k;$ the latter sequence definitely converges to M. The above should be enough to write a real proof, either the epsilon/delta-way or using standard facts about (Cauchy) sequences.

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