MHB Convergence of Alternating Series and Estimation of Absolute Error

[mathmari]

Hey!

I want to determine the value of $n\in \mathbb{N}$ such that the absolute error $$\left |\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)\cdot k!}-\sum_{k=0}^{n}\frac{(-1)^k}{(2k+1)\cdot k!}\right | =\left |\sum_{k=n+1}^{\infty}\frac{(-1)^k}{(2k+1)\cdot k!}\right |$$ is less than $10^{-6}$. Do I have to find an upper bound for the error? But how? I tried the following:
$$\left |\sum_{k=n+1}^{\infty}\frac{(-1)^k}{(2k+1)\cdot k!}\right |\leq \sum_{k=n+1}^{\infty}\left |\frac{(-1)^k}{(2k+1)\cdot k!}\right |=\sum_{k=n+1}^{\infty}\frac{1}{(2k+1)\cdot k!}\leq \frac{1}{(2(n+1)+1)\cdot (n+1)!}$$ But this tends to infinity, doesn't it? We have here an alternating series. The sequence $a_k:=\frac{1}{(2k+1)\cdot k!}$ is decreasing. That means that $a_{n+1}$ is greater than every other term of the sum. Is it also bigger than the whole sum? (Wondering)

 

[I like Serena]

We have here an alternating series. The sequence $a_k:=\frac{1}{(2k+1)\cdot k!}$ is decreasing. That means that $a_{n+1}$ is greater than every other term of the sum. Is it also bigger than the whole sum? (Wondering)

Hey mathmari! (Smile)

The term $a_{k+1}$ is not bigger then the whole sum, but it is bigger than the remaining error of the series up to $a_k$.
See for instance the Formulation of the Alternating series test.

 

[mathmari]

The term $a_{k+1}$ is not bigger then the whole sum, but it is bigger than the remaining error of the series up to $a_k$.
See for instance the Formulation of the Alternating series test.

Ah ok! So, we have the following:

\begin{align*}\left |\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)\cdot k!}-\sum_{k=0}^{n}\frac{(-1)^k}{(2k+1)\cdot k!}\right | & \leq \frac{1}{(2(n+1)+1)\cdot (n+1)!} \\ & =\frac{1}{(2n+2+1)\cdot (n+1)!} \\ & =\frac{1}{(2n+3)\cdot (n+1)!}\end{align*}

Since the error should be less that $10^{-6}$ it must hold \begin{equation*}\frac{1}{(2n+3)\cdot (n+1)!}<10^{-6}\Rightarrow (2n+3)\cdot (n+1)!>10^6\end{equation*} How can we solve this inequality for $n$ ? (Wondering)

 

[I like Serena]

How about calculating a number of subsequent values? We should get there pretty quick.
We can choose to keep only 1 or 2 significant digits and round down. (Thinking)

 

[mathmari]

How about calculating a number of subsequent values? We should get there pretty quick.
We can choose to keep only 1 or 2 significant digits and round down. (Thinking)

I substituted some values of $n$. The smallest $n$ that satisfies that inequality is $n=8$.

 

[I like Serena]

Sounds about right. (Nod)

 

[mathmari]

Sounds about right. (Nod)

(Yes)

One last question to clarify... We have the infinite sum $\displaystyle{\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)\cdot k!}}$. Is the $n$-th partial sum $\displaystyle{\sum_{k=0}^{n}\frac{(-1)^k}{(2k+1)\cdot k!}}$, i.e. till $k=n$, or $\displaystyle{\sum_{k=0}^{n-1}\frac{(-1)^k}{(2k+1)\cdot k!}}$, i.e., that the sum contains $n$ terms?

(Wondering)

 

[I like Serena]

It seems that is not clearly defined.
For instance wiki starts with '... sums of the n first terms of the series, which are called the nth partial sums of the series.'
But in the next section it says that the kth partial sum is from 0 to k.

My view: if we have a sequence of partial sums $s_0, s_1, ..., s_n, ...$, then $s_n$ represents the nth partial sum. (Thinking)

 

[mathmari]

It seems that is not clearly defined.
For instance wiki starts with '... sums of the n first terms of the series, which are called the nth partial sums of the series.'
But in the next section it says that the kth partial sum is from 0 to k.

My view: if we have a sequence of partial sums $s_0, s_1, ..., s_n, ...$, then $s_n$ represents the nth partial sum. (Thinking)

Ok! Thank you very much! (Handshake)