If n-1983- then the value of expression 1-log2 n-1-log3 n-1-log4 n-1-log1983n is equal to

Log_n 2+log_n3+… log_n 1983 =log_n[2.3.4…1983]=log_n 1983! =log_n n=1 ∵ 1983!=n ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Fundamental Laws of Logarithms

If n=1983!, t...

Question

If n=1983!, then the value of expression $\frac{1}{l o g_{2} n} + \frac{1}{l o g_{3} n} + \frac{1}{l o g_{4} n} + \dots \frac{1}{l o g_{1983} n}$ is equal to

-1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

Similar questions

n = (2017)!

\frac{1}{{log}_{2} n} + \frac{1}{{log}_{3} n} + \frac{1}{{log}_{4} n} + . . . . + \frac{1}{{log}_{2017} n}

\frac{1}{l o g_{2} n} + \frac{1}{l o g_{3} n} + \frac{1}{l o g_{4} n} + \dots \frac{1}{l o g_{1983} n}

\frac{1}{{log}_{2} N} + \frac{1}{{log}_{3} N} + \frac{1}{{log}_{4} N} + . . . . . . \frac{1}{{log}_{100} N}

(N \neq 1)

\frac{1}{{log}_{2} . n} + \frac{1}{{log}_{3} n} + \frac{1}{{log}_{4} n} + \dots + \frac{1}{{log}_{1983} n} .

N = n! (n \in N, n > 2)

lim N \to \infty [{({log}_{2} N)}^{- 1} + {({log}_{3} N)}^{- 1} + . . . . . {({log}_{n} N)}^{- 1}]

Join BYJU'S Learning Program