Question

Find the sum of $\dfrac{3}{1!}+\dfrac{5}{2!}+\dfrac{9}{3!}+\dfrac{15}{4!}+\dfrac{23}{5!}+.............\infty$
1. $4e-1$
2. $4e-3$
3. $3e+2$
4. $3e+4$

Answer 1

In these types of problem, firstly we have to observe the given series and find the ${{n}^{th}}$ term of given series and after that we have to simplify that term and perform the summation, after summation we have to check and substitute the series as sum of Euler’s number and then after simplifying It further, you will get your required answer.

Complete step by step answer:
A series can simply be defined as the sum of the various numbers, or elements of a sequence. The series can be finite or infinite depending on the sequence whether it is finite or infinite.
Sequence can be defined as the set of the elements that follow a certain pattern whereas series can be defined as the sum of elements of the given sequence. The finite series are series where the numbers are ending and infinite series are the series where the numbers are never ending.
Types of series are as follows:
Geometric Series
Harmonic Series
Power Series
Alternating Series
Exponent Series
A geometric series can be defined as a series with a constant ratio between successive terms.
A harmonic series can be defined as the series that contains the sum of terms that are the reciprocal of the arithmetic series terms.
Power series can be defined as the series that can be thought of as a polynomial with an infinite number of terms.
As we are given in the question: $\dfrac{3}{1!}+\dfrac{5}{2!}+\dfrac{9}{3!}+\dfrac{15}{4!}+\dfrac{23}{5!}+.............\infty$
As, $e$ is a real number which is called Euler's number. We can represent it as the sum of infinite numbers, such as:
$e=1+\dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+..............$
By observing the given series, the ${{n}^{th}}$ term of the series can be given as:
${{T}_{n}}=\dfrac{{{n}^{2}}-n+3}{n!}$
Now we will decompose our term as:
$\Rightarrow {{T}_{n}}=\dfrac{{{n}^{2}}}{n!}-\dfrac{n}{n!}+\dfrac{3}{n!}$
Now we will do the summation of above term:
$\Rightarrow {{T}_{n}}=\sum{\dfrac{{{n}^{2}}}{n!}}-\sum{\dfrac{n}{n!}}+\sum{\dfrac{3}{n!}}$
Substitute the values of $n$ from $1,2,......\infty$
$\Rightarrow {{T}_{n}}=\dfrac{1}{1!}+\dfrac{{{2}^{2}}}{2!}+\dfrac{{{3}^{2}}}{3!}+........\infty -\left[ \dfrac{1}{1!}+\dfrac{2}{2!}+\dfrac{3}{3!}+.......\infty \right]+3\left[ \dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{3!}+.........\infty \right]$
It can be simplify as follows:
$\Rightarrow {{T}_{n}}=2e-e+3(e-1)$
$\Rightarrow {{T}_{n}}=e+3e-3$
$\Rightarrow {{T}_{n}}=4e-3$
Since, the sum of series can be given as: $4e-3$

So, the correct answer is “Option 2”.

Note: Euler’s number is an irrational mathematical constant that forms the base of all natural logarithms. Euler’s number is used in the most practical sense for working with radioactive decay. In mathematics, it is a crucially important tool that allows mathematicians to convert complex numbers to more usable form.