Let $(b_n)_{n \geq 0}$ a sequence of complex numbers such that $\sum_{n=0}^{\infty} |b_n| < \infty$. Let's suppose also that $$ \sum_{n=0}^{\infty} \frac{b_n}{k^n}=0 $$ For each integer $k \geq 2$. I want to show that $b_n=0$ for all $n$.

My attempt:

If $(z_n)$ is any sequence of complex numbers we know that $z_n \rightarrow z$ if and only if $Re (z_n) \rightarrow Re (z)$ and $Im (z_n) \rightarrow Im (z)$. Therefore, it's enough to show the result in the case when $b_n \in \mathbb{R}$. Let's assume then that $b_n \in \mathbb{R}$. Since $\sum_{n=0}^{\infty} |b_n| < \infty$ it follows that the series formed from its positive terms and the series formed from its negative terms both converge. We can then write $b_n=b_n^+ - b_n^{-}$ for all $n$, and $(b_n^+), (b_n^{-})$ are two sequences of non-negative numbers such that their respective series converge.

Since $b_n=b_n^+ - b_n^{-}$, we have then that: $$ \left| \sum_{j=0}^{n} \frac{b_j}{k^n} \right| \geq \left| \left| \sum_{j=0}^{n} \frac{b_j^{+}}{k^n} \right| -\left| \sum_{j=0}^{n} \frac{b_j^{-}}{k^n} \right| \right| \geq0 $$ Therefore, since the sequence of the left converges to $0$ and $\sum_{j=0}^{n} \frac{b_j^{+}}{k^n} ,\sum_{j=0}^{n} \frac{b_j^{-}}{k^n} $ both converge, we have that $$ \left| \left| \sum_{j=0}^{n} \frac{b_j^{+}}{k^n} \right| -\left| \sum_{j=0}^{n} \frac{b_j^{-}}{k^n} \right| \right| \rightarrow 0 \textrm{ as } n \rightarrow \infty $$

Which implies that $$ \sum_{j=0}^{\infty} \frac{b_j^{+}}{k^n} =\sum_{j=0}^{\infty} \frac{b_j^{-}}{k^n} $$ Of course, from here I'd love to conclude that there exist $N \in \mathbb{N}$ such that $b_j^{+}=b_j^{-}$ for $n \geq N$ which would basically finish the proof. However, this doesn't follow necessarily from the last iquality since we can have two sequences of non-negative numbers with the same sum but which are not eventually equal as I suggested above. In other words, in general $$ \sum_{j=0}^{\infty}a_j =\sum_{j=0}^{\infty} b_j $$ doesn't imply that the sequences are almost equal. As you can see I am missing something.

How can I conclude/fix this issue?

In advance thank you very much.

Edit: I am looking to solve this exercise using the basics about sequences and series (or even known tools of real analysis). There is probably an easier solution using more advanced tools of complex analysis but I haven't seen them yet.

  • $\begingroup$ Just take $b_n=\frac{1}{r^n} ; r>=1$ and your series converges- $\endgroup$
    – trula
    Feb 24 at 21:28
  • $\begingroup$ @trula That sequence doesn't fulfill the required condition? $\endgroup$
    – Bruno B
    Feb 24 at 21:39
  • 3
    $\begingroup$ Take $f(z)=\sum b_nz^n, |z|<1$ which is analytic in the unit disc because $\sum |b_n| <\infty$; then $f(1/k)=0, k \ge 0$ and $1/k$ has a limit point at $0$ so one concludes that $f=0$ $\endgroup$
    – Conrad
    Feb 24 at 21:54
  • $\begingroup$ @Conrad you're probably right! But I was looking to solve it using basic notions of sequences and series. I just started a complex analysis course so I guess the purpose of the exercise is to solve it with the current tools. I should have probably said that in the question, my bad! Thank you very much though :) $\endgroup$
    – ferolimen
    Feb 24 at 23:19
  • 1
    $\begingroup$ then you would have to sort of mimic the proof of the result above and use induction; show using that $\sum |b_n|$ finite that if $S_k=\sum_{n=1}^{\infty} \frac{b_n}{k^n}$ then $S_k \to 0, k \to \infty$ (since $1/2^k,...$ decrease fast) hence $b_0=0$ since $b_0+S_k=0$; then by induction, let $n \ge 1$ the first index for which $b_m \ne 0$ and divide by $1/m^k$ every sum and get $b_m+\sum_{n \ge 1} b_{m+n}c_n^k =0, c_n=(m+n)/n >\delta=1+1/m>1$ and again show that as $k \to \infty$ the sum $S_{k,m}=\sum_{n \ge 1} b_{m+n}c_n^k \to 0$ which implies $b_m=0$ and you are done $\endgroup$
    – Conrad
    Feb 24 at 23:38

1 Answer 1


It suffices to assume that the sequence $(b_n)$ is bounded (which is the case if $\sum_{n=0}^{\infty} |b_n| $ is convergent).

So let $(b_n)$ be a (real or complex) sequence with $|b_n| \le M$ for all $n$, and with $\sum_{n=0}^{\infty} \frac{b_n}{k^n}=0 $ for all integers $k \ge 2$.

Assume that not all $b_n$ are zero, and let $m$ be the smallest index with $b_m \ne 0$. Then $$ 0 = \sum_{n=0}^{\infty} \frac{b_n}{k^n} = \frac{b_m}{k^m} + \sum_{n=m+1}^\infty \frac{b_n}{k^n} $$ and therefore $$ 0 < \left|b_m \right| = \left|\sum_{n=m+1}^\infty \frac{b_n}{k^{n-m}} \right| \le M \sum_{n=m+1}^\infty \frac{1}{k^{n-m}} = M \frac{1}{k-1} $$ for all integers $k \ge 2$. That is not possible because the right-hand side converges to zero for $k \to \infty$.


  • It is correct that the statement can be reduced to the case of real sequences, but that reduction is not necessary: The above argument uses only the triangle inequality (which holds for complex numbers as well) and the formula for the geometric series.

  • Which means that the same statement holds for sequences $(b_n)$ in any normed space over $\Bbb R$ or $\Bbb C$, with the identical proof.

  • The proof also shows that the same conclusion holds if $\sum_{n=0}^{\infty} \frac{b_n}{k^n}=0$ for arbitrarily large (real or complex) numbers $k$. That is not surprising if one knows about the identity theorem for holomorphic functions, which can be applied to $f(z) = \sum_{n=0}^{\infty} b_n z^n$ (as mentioned in the comments).

  • $\begingroup$ Your first inequality doesn’t make sense to me. $|b_m|=|\sum_{n>m}b_n/k^{n-m}|$ because..? Seems to me that we should include the lower tail $\sum_{n<m}$ too $\endgroup$
    – FShrike
    Feb 25 at 12:32
  • $\begingroup$ @FShrike: $0 = \sum_{n=0}^{\infty} \frac{b_n}{k^n} = \frac{b_m}{k^m} + \sum_{n> m} \frac{b_n}{k^n}$ because $b_m$ is the first nonzero element in the sequence. $\endgroup$
    – Martin R
    Feb 25 at 12:34
  • $\begingroup$ Yeah, that was stupid. $\endgroup$
    – FShrike
    Feb 25 at 12:37
  • $\begingroup$ @MartinR What an elegant proof! Thank you, and also thanks for the further remarks! As for the comment you wrote above: you're right, I don't know why I wrote "null" instead of "zero". $\endgroup$
    – ferolimen
    Feb 25 at 18:27

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