We can factor this out of the series:
$$
\frac{1}{3}\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{9}\right)
$$
Now, we can see that the series inside the parentheses is a sum of the reciprocals of the first 9 positive integers. We can write this
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