This question was actually derived from a time complexity recurrence relation question. Please also explain how is this a harmonic series.?$$\frac 1{\log (n)- i}$$
Asked
Active
Viewed 253 times
2
-
Let $N = \log(n)$ (we assume it's an integer here; it's fine for complexity computations). You are summing all numbers from $1/1$ to $1/(N-1)$, but in the opposite order: $\frac{1}{N-1} + \frac{1}{N-2} + \ldots + \frac{1}{2 } +\frac{1}{1}$. Try to do a substitution $j = \log(n) - i$ and compute the limits on $j$ from the limits on $i$. – Winther Mar 15 '19 at 08:50
-
please can you explain how can solve this through arithmetic progression – S.Ohanzee Mar 15 '19 at 09:12
-
For "time complexity" you only need to know how fast it grows asymptotically https://en.wikipedia.org/wiki/Harmonic_series_(mathematics)#Rate_of_divergence – Winther Mar 15 '19 at 09:16
-
how ? explain this plz – S.Ohanzee Mar 15 '19 at 09:35
1 Answers
0
Is $n$ supposed to be $k$? Otherwise, $\log (n)$ can be replaced by some constant $a$. If $a=k$ this is a partial sum of the harmonic series. For any $a$ this is a partial sum of the reciprocals of an arithmetic progression, i.e. of a harmonic progression. There is no closed form formula for partial sums of harmonic progressions (or even of harmonic series), but there are very good asymptotic approximations, see Is there a partial sum formula for the Harmonic Series? and sum of harmonic progression?
Conifold
- 11,756
-
-
-
by solve means my teacher informed me that harmonic progression cannot be solved until and unless you converge it into arithmetic function – S.Ohanzee Mar 18 '19 at 12:03
-
@S.Ohanzee To "solve" a progression usually means finding a closed formula for its general term, but you already have that. In the current version of OP there is no mention of the sum, so I do not know what the question is. Does "converge" mean that you want an asymptotic formula for the sum? – Conifold Mar 19 '19 at 03:44