Prove that every subsequence of a summable sequence with non-negative terms is summable.

real-analysis

I was currently trying to prove the following statement:

"Prove that every subsequence of a summable sequence with non-negative terms is summable"

I wanted to use the Convergence Criterion for Monotone Sequences to prove this statement. Since the terms of the summable sequence are non-negative, we also know that the terms of the subsequence are non-negative. Therefore, we know that the sequence of partial sums of the subsequence is increasing. Now I am struggling to show that the subsequence is bounded from above.

Can I simply say that the sequence of partial sums of the subsequence is in fact a subsequence of the partial sums of the sequence? And if so why?

I was thinking about using the following theorems:

  • Every convergent sequence is bounded.
  • Every subsequence of a convergent sequence is convergent and converges to the limit of the sequence.

I was hoping somebody could help me with proving this statement.

Best Answer

You cannot simply say that the partial sums of the subsequence is a subsequence of the partial sums of the sequence. Consider the sum of $a_n=\frac{1}{n(n+1)}$

$$1=\sum_{n=1}^\infty\frac{1}{n(n+1)}=\frac{1}{2}+\sum_{n=2}^\infty\frac{1}{n(n+1)}$$

Thus, if $b_n$ is any subsequence of $a_n$ without $a_1$ we have

$$\sum_{n=1}^N b_n<\sum_{n=2}^\infty\frac{1}{n(n+1)}=\frac{1}{2}$$

which is less than any partial sum of $a_n$. What you can say though is

$$\sum_{k=1}^N a_{n_k}\leq \sum_{k=1}^{n_N} a_{n}\leq \sum_{k=1}^{\infty} a_{n}=L<\infty$$


EDIT: With more explanation

$$\sum_{k=1}^N a_{n_k}\leq \sum_{k=1}^{n_N} a_{n}$$

This is true since the sum on the left is the partial sums of the subsequence. Note that the sum ends at the subsequence term $a_{n_N}$, which has index $n_N$. The sum on the right is the partial sums of every term in the sequence with index less than $n_N$.

$$\sum_{k=1}^{n_N} a_{n}\leq \sum_{k=1}^{\infty} a_{n}=L$$

Here, the sum on the left is the finite sum up to the $n_N$ term. The sum on the right is the infinite sum. By assumption, the infinite sum converges.


Since $\sum_{k=1}^N a_{n_k}$ is an increasing sequence in $N$ and bounded above, it must have a limit.