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This question already has an answer here:

\begin{align} T(0) & = 0 \\ T(n) & = T(n-1) + \dfrac{1}{n} \end{align} solve the recurrence relation

My work so far:

\begin{align} T(1) & = 1 \\ T(2) & = 1 + \dfrac{1}{2} \\ T(3) & = 1 + \dfrac{1}{2} + \dfrac{1}{3} \\ &\vdots \end{align}

this is the harmonic series, which diverges.

What is the solution to the recurrence relation?

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marked as duplicate by Bookend, Brian Fitzpatrick, Lord_Farin, quid, Arctic Char Feb 21 '15 at 22:49

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

The actual recurrence relation is the first line of the question. If this is the whole problem, you’re probably expected to prove by induction that $T(n)=H_n$, the $n$-th harmonic number, for $n\ge 1$. – Brian M. Scott Dec 30 '12 at 11:20

The harmonic series diverges when you take the sum to infinity, but this is just the sum to n, which can be calculated. What I'm basically saying is that I think the harmonic series is the right approach.

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