Question

Guys, please help with this question. Given that S = {S_n;S_n = 1+\sum_{i=1}^{n^2} (\frac{1}{i!}), n = 1,2... }

Show that 3\in U_s

Answer 1

I see you've tried to present this problem properly using the Equation Editor, but for some reason did not succeed. Could you possibly draw this equation in the Draw utility, below?

Answer 2

\[ S = {S_n;S_n = 1+\sum_{i=1}^{n^2} (\frac{1}{i!}), n = 1,2... } \]

Answer 3

Show that \(3\in U_s\)

Answer 4

some confusion here clearly
there is no \(U_S\) in the question

Answer 5

I am confused on this as well as i cannot identify what U_s stands for

Answer 6

I think they want you show \(3 \not\in S\)

Answer 7

Some form of union?

Answer 8

Looks like \(S_n=\exp_{n^2}(1)=\displaystyle\frac{e\Gamma(n^2+1,1)}{\Gamma(n^2+1)}\) where \(\exp_n(x)=\displaystyle\sum_{i=0}^n\frac{x^i}{i!}\) denotes the exponential sum function http://mathworld.wolfram.com/ExponentialSumFunction.html
and \(\Gamma(n,x)\) and \(\Gamma(n)\) the (incomplete) gamma functions
http://mathworld.wolfram.com/IncompleteGammaFunction.html
http://mathworld.wolfram.com/GammaFunction.html
Not that that gives any more information... Are you sure you've never seen this notation defined or used anywhere else?

Answer 9

It is about showing that 3 is an element of upper bound of the set S

Answer 10

\[\begin{align}1+\color{blue}{\sum_{i=1}^{\infty} (\frac{1}{i!})} &= 1+\color{blue}{\dfrac{1}{1} + \dfrac{1}{1\cdot 2} + \dfrac{1}{1\cdot 2\cdot 3} + \cdots}\\~\\
&\lt 1 + \color{blue}{\dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{2\cdot 2} +\dfrac{1}{2\cdot 2\cdot 2}+ \cdots}\\~\\
&= 1+ \color{blue}{2}

\end{align}\]

Since the sequence \(S_n\) is increasing, all the terms in the sequence are less than 3.

Answer 11

@ganeshie8 Okay..Thanks for the insight....I get the 1st Line as this appears to be order of factorials...However i am still in dark as regards the second line.... can u please explain the second line?

Answer 12

Notice that


\(1\cdot 2 \cdot 3 \gt 2\cdot 2\)

Answer 13

That means

\(\dfrac{1}{1\cdot 2 \cdot 3} \lt \dfrac{1}{2\cdot 2}\)

Answer 14

\[\begin{align}1+\color{blue}{\sum_{i=1}^{\infty} (\frac{1}{i!})} &= 1+\color{blue}{\dfrac{1}{1} + \dfrac{1}{1\cdot 2} + \dfrac{1}{1\cdot 2\cdot 3} + \dfrac{1}{1\cdot 2\cdot 3\cdot 4} + \cdots}\\~\\
&\lt 1 + \color{blue}{\dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{2\cdot 2} +\dfrac{1}{2\cdot 2\cdot 2}+ \cdots}\\~\\
&= 1+ \color{blue}{2}

\end{align}\]

Since the sequence \(S_n\) is increasing, all the terms in the sequence are less than 3.

Answer 15

thanks