Algorithm substitution method

Question

How to find  log n base2+ log n base 3+ log n base4+........log n base n?

Comments

How to solve 1/log2 +1/log4+1/log6...+1 /logn?

Answer 1

= log₁n + log₂n + log₃n +log₄n .............log_nn

= logn/log1 + logn /log2 +logn/log3 +logn/log4 .......... logn/logn (By using property of log )

= logn ( 1/log1 +1/log2 + 1/log3 +1/log4............1/logn ) 

= O(logn)

Comments

Log1=0, so 1/log1= infinity so how can we ignore the infinty in last step although it is decreasing gp

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we can not ignore infinity in the last step ...and by this, we can't decide its asymptotic value

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@Anil JI , Can u explain what's wrong in his solution except log n / log 1 part because it is not given in the question and also it is undefined..

Answer 2

Answer is 1 for n=1 and infinity for all other values.

This can be done by substituting 1 and 2,3,4..in place of n.

For n=1, log 1 base 1 is 1 while for n=2, 3.. Log2 base 1 is infinity

Comments

Did not understand plzz elaborate .i want to find asymtotic value of this

Answer 3

f(n)=logn{1/log2+1/log3+1/log4+1/log5+1/log6+1/log7+1/log/8+1/log9+1/log10+1/log11+.........1/logn} (base is 10)

let sum of first nine term be k where k is some constant and sum of remaining term will be <2(less than 2,let it's x))

f(n)=logn{k+x}

f(n)=klogn+xlogn

so,f(n)=O(logn)