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\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 Yes, Brian Fitzpatrick, Lord_Farin, quid, John Feb 21 at 22:49

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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

1 Answer 1

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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