The value of $\dfrac{1}{{{\log }_{3}}e}+\dfrac{1}{{{\log }_{3}}{{e}^{2}}}+\dfrac{1}{{{\log }_{3}}{{e}^{4}}}+...$ upto infinite terms is:
(A) ${{\log }_{e}}9$
(B) 0
(C) 1
(D) ${{\log }_{e}}3$

Question

The value of $\dfrac{1}{{{\log }_{3}}e}+\dfrac{1}{{{\log }_{3}}{{e}^{2}}}+\dfrac{1}{{{\log }_{3}}{{e}^{4}}}+...$ upto infinite terms is:(A) ${{\log }_{e}}9$(B) 0(C) 1(D) ${{\log }_{e}}3$

Vedantu Content Team · Accepted Answer

Hint: We have to change the base of the logarithm so that the terms will get simplified and form a geometric progression. Then apply the formula of sum of infinite terms of a G.P. which is $S=\dfrac{a}{1-r}$, where $a$ is the first term and $r$ is the common ratio of the G.P.

Complete step-by-step answer:
There are different types of series but mainly we have to study three of them namely Arithmetic progression (A.P.), Geometric Progression (G.P.) and Harmonic Progression (H.P.). In this question geometric progression will be used.

In Mathematics, a G.P. also known as geometric sequence is a sequence of numbers where each term after its predecessor is obtained by multiplying the previous term with a fixed non-zero number known as the common ratio of the G.P.

For example: \[2,\text{ }4,\text{ }8,\text{ }16,\text{ }32,\text{ }.........\] is a G.P. with common ratio $2$.

Generally, G.P. is represented by $a,\text{ }a{{r}^{2}},\text{ }a{{r}^{3}},\text{ }\ldots \ldots \ldots $ where \[a\] is the first term and \[r\] is the common ratio.

Now, we have been given:

${{S}_{\infty }}=\text{Sum of infinite series}=\dfrac{1}{{{\log }_{3}}e}+\dfrac{1}{{{\log }_{3}}{{e}^{2}}}+\dfrac{1}{{{\log }_{3}}{{e}^{4}}}+.........\infty $

We can change the base of the logarithm by using the following formula:

\[{{\log }_{m}}n=\dfrac{1}{{{\log }_{n}}m}\]

For example; \[{{\log }_{3}}e=\dfrac{1}{{{\log }_{e}}3}\].

$\therefore {{S}_{\infty }}={{\log }_{e}}3+{{\log }_{{{e}^{2}}}}3+{{\log }_{{{e}^{3}}}}3+.........\infty $
Now, using the following property of logarithm: ${{\log }_{{{m}^{a}}}}n=\dfrac{1}{a}{{\log }_{m}}n$, we get

$\therefore {{S}_{\infty }}={{\log }_{e}}3+\dfrac{1}{2}{{\log }_{e}}3+\dfrac{1}{3}{{\log }_{e}}3+.........\infty $

We can clearly see that this a G.P. with the first term $a={{\log }_{e}}3$ and $r=\dfrac{1}{2}$.

$\therefore {{S}_{\infty }}=\dfrac{a}{1-r}=\dfrac{{{\log }_{e}}3}{1-\dfrac{1}{2}}=2{{\log }_{e}}3$

Now, using the rule that: \[a\times {{\log }_{m}}n={{\log }_{n}}({{n}^{a}})\], we get

${{S}_{\infty }}={{\log }_{e}}({{3}^{2}})={{\log }_{e}}9$

Hence the correct option is (A).

Note: Always try to convert the question in simplified form just as we did here. We first simplified the terms of the given sequence by converting the fractions into simple terms and then applied the formula for summation of infinite terms of a G.P. Here, we have to sum infinite terms so the formula for summation of $n$ terms cannot be used.

The value of $\dfrac{1}{{{\log }_{3}}e}+\dfrac{1}{{{\log }_{3}}{{e}^{2}}}+\dfrac{1}{{{\log }_{3}}{{e}^{4}}}+...$ upto infinite terms is:
(A) ${{\log }_{e}}9$
(B) 0
(C) 1
(D) ${{\log }_{e}}3$

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The value of $\dfrac{1}{{{\log }_{3}}e}+\dfrac{1}{{{\log }_{3}}{{e}^{2}}}+\dfrac{1}{{{\log }_{3}}{{e}^{4}}}+...$ upto infinite terms is:(A) ${{\log }_{e}}9$(B) 0(C) 1(D) ${{\log }_{e}}3$

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NCERT Book Solutions

Reference Book Solutions

Question Papers

Exams

Quick Links

The value of $\dfrac{1}{{{\log }_{3}}e}+\dfrac{1}{{{\log }_{3}}{{e}^{2}}}+\dfrac{1}{{{\log }_{3}}{{e}^{4}}}+...$ upto infinite terms is:
(A) ${{\log }_{e}}9$
(B) 0
(C) 1
(D) ${{\log }_{e}}3$