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What is the complexity of the follwoing recurrence? $$T(n) = T(n-1) + 1/n$$

I highly suspect the answer to be $O(1)$, because your work reduces by $1$ each time, so by the $n$th time it would be $T(n-n) = T(0)$ and your initial task reduces to 0.

Soham Chowdhury
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Olórin
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2 Answers2

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By unfolding $$T(n) = T(n-1) + \frac{1}{n} = T(n-2) + \frac{1}{n} + \frac{1}{n-1} = \dots=T(0) + \sum_{k=1}^{n} \frac{1}{k}$$ Now we can easily approximate the sum on the RHS using that $$\sum_{k=1}^{n}\frac{1}{k} \le 1 + \int_{1}^{n}\frac{1}{x} dx = 1+ \log{n} - \log{1} = 1+ \log{n}$$ Therefore $T(n) \equiv O(\log{n})$

0xdeadcode
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Edit

This answer assumed that the OP was looking for the time-complexity of an algorithm that would evaluate the recurrence, so it's probably wrong.

Whatever you do, you have to iterate $n$ times, whatever the base case/starting value. So as $n$ grows without bound, the number of operations also grows linearly with $n$ - implying that the recurrence is $O(n)$.

Look at it this way: To calculate $T(n)$, you need $T(n-1)$, which in turn depends on $T(n-2)$, and so on all way till $T(0)$ (or whatever the lowest allowed value of $n$ is).

Each time, you calculate $\frac{1}{k}$ and add it to the next lower value of $T(k)$, doing $O(1)$ work each time, $n$ times - which adds up to $O(n)$ complexity.

In fact, you can easily represent $T(n)$ in general as

$$ T(n) = T(0) + \sum^n_{k=1}{\frac{1}{k}} $$

(which, if you're curious, is equal to $T(0) +H_n$, where $H_n$ is the $n$th harmonic number.)

Soham Chowdhury
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