Question

If \( S = \sum_{n=0}^{\infty} \frac{(\log x)^{2n}}{(2n)!} \), then \( S \) is equal to

• A. \( x + x^{-1} \)
• B. \( x - x^{-1} \)
• C. \( \dfrac{1}{2}\left(x + x^{-1}\right) \)
• D. None of these

Hint:

Whenever a series contains only even powers like \( \dfrac{t^{2n}}{(2n)!} \), think of the standard expansion of \( \cosh t \).

Answer 1

The Correct Option is C

Step 1: Write the given series clearly.
We are given \[ S = \sum_{n=0}^{\infty} \frac{(\log x)^{2n}}{(2n)!} \] This is an infinite series involving only even powers of \( \log x \)

Step 2: Recall the standard expansion.
We know that the Maclaurin series expansion of \(\cosh t\) is \[ \cosh t = \sum_{n=0}^{\infty} \frac{t^{2n}}{(2n)!} \] Comparing this with the given series, we can take \[ t = \log x \]

Step 3: Identify the series.
So, the given sum becomes \[ S = \cosh(\log x) \] Now we use the standard identity for hyperbolic cosine

Step 4: Use the formula for \(\cosh y\).
We know that \[ \cosh y = \frac{e^y + e^{-y}}{2} \] Putting \( y = \log x \), we get \[ \cosh(\log x) = \frac{e^{\log x} + e^{-\log x}}{2} \]

Step 5: Simplify the exponential terms.
Using the properties of logarithms and exponentials, \[ e^{\log x} = x \] and \[ e^{-\log x} = \frac{1}{e^{\log x}} = \frac{1}{x} = x^{-1} \] Therefore, \[ S = \frac{x + x^{-1}}{2} \]

Step 6: Match with the given options.
The obtained result is \[ S = \frac{1}{2}\left(x + x^{-1}\right) \] This matches option \((3)\)

Step 7: Final conclusion.
Hence, the correct value of the given infinite series is \[ \boxed{\frac{1}{2}\left(x + x^{-1}\right)} \] So, the correct answer is (C)