Stuck in Recurrence Relation in Algorithm

Question

How to find log of the first n natural numbers?

$T(n)= \log 1+ \log 2+ \log 3+\ldots + \log n$  // How to proceed from here?

Comments

log ($ 1\times2 \times3\times4\times......\times n$) = log(n!) = log(nⁿ) = O(nlogn)

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i think n! can be maped to exponential ....n^{n  ...}

so O(nlogn)

Answer 1

you can write  log( 1 *2 *3 * ------------------n)

log( n ! )   answer    product of 1st n natural number = n ! 

other one :  1 + 2+ 3+ -------------+ n  = n(n+1) /2 

same way log 1 + log 2+ -------- log n = $\frac{logn(logn +1 )}{2}$   {  but it is not true  } 

Comments

I guess log1+log2+log3+... isn't following an AP series..so 2nd one can't be applied.

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@MiNiPanda  yes i think too because i cross checked by taking values it didnot hold  

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Yes..true

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@sumit goyal 1 Anu007 how  you you write log1+ log2+ log3+⋯+ logn as  log( 1 *2 *3 * ------------------n) as 

log( n ! ) 

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hi bhai   by log property   log a + log b = log ( a.b)

so if you have , log a + log b + log c +------- + log n  then it  is equal to

log(a . b . c -------  . n )

and  n!  = 1.2.3.4............ n   @iarnav

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@sumit goyal 1 Thanks Bhaijaan :)