Error accumulation in an approximating numerical algorithm for $y_n =\int_{0}^{1} \frac{x^n}{x+10} dx $ Consider the problem of calculating the integral
$$y_n =\int_{0}^{1} \dfrac{x^n}{x+10} \mathrm{d}x $$
for a positive integer $n$.
Observe that $$y_n + 10y_{n-1} = \int_{0}^{1} \dfrac{x^n +10x^{n-1}}{x+10} \mathrm{d}x  = \int_{0}^{1} x^{n-1}\mathrm{d}x = \dfrac{1}{n}$$
and that using this relationship in a forward recursion leads to a numerically unstable procedure.
$(a)$ Derive a formula for approximately computing these integrals based on evaluating $y_{n-1}$ given $y_n$.
$(b)$ Show that for any given value $\epsilon > 0$ and positive integer $n_0$, there exists an integer $n_1 \geq n_0$ such that taking $y_{n_1} = 0$ as a starting value will produce integral evaluations $y_n$ with an absolute error smaller than $\epsilon$ for all  $0 < n \leqslant n_0$. 
$(c)$ Explain why your algorithm is stable.
Here is what I have so far,
for part(a)
$$y_{n-1} = \dfrac{1}{10} \left(\dfrac{1}{n} - y_n\right)$$
and for part $(c)$
The algorithm is stable because the magnitude of roundoff errors gets divided by 10 each time the recursion is applied.
I really don't know how to start on the proof for part $(b)$, any hints and help would be appreciated.
 A: When $y_{n_1} = 0$, $y_{n_1-1} = \dfrac{1}{10n_1} + \alpha_1$, where $\alpha_n$ is the value of the roundoff error at each step (not necessarily the same for each step.)  Plugging this back into the formula:
$y_{n_1-2} = \dfrac{1}{10} \left(\dfrac{1}{n_1} - y_{n_1 - 1}\right)$
we get: 
$y_{n_1-2} = \dfrac{1}{10} \left(\dfrac{1}{n_1 - 1} - \left(\dfrac{1}{10n_1} + \alpha_1\right)\right) + \alpha_2 = 
\dfrac{1}{10(n_1 - 1)} - \dfrac{1}{100n_1} - \dfrac{\alpha_1}{100} + \alpha_2$
Repeating, we get:
$y_{n_1-3} = \dfrac{1}{10} \left(\dfrac{1}{n_1 - 2} - y_{n_1 - 2}\right) + \alpha_3 = 
\dfrac{1}{10(n_1 - 2)} - \dfrac{1}{100(n_1 - 1)} - \dfrac{1}{1000n_1} - \dfrac{\alpha_1}{1000} -\dfrac{\alpha_2}{100} + \alpha_3.$
As you say, the magnitude of the roundoff error from the previous step is divided by 10 each time the recursion is applied, and forms a geometric series.  For a given $n_1$, the worst case magnitude of the error at any $n_0 \geq 1, n_1 \geq 2, n_1 > n_0$ is bounded by:
$$|\alpha_{max}|\sum_{k=0}^{n_1 - n_0 - 1} \left(\dfrac{1}{10}\right)^k= |\alpha|\dfrac{1 - {\left(\dfrac{1}{10}\right)}^{n_1 - n_0}}{1 - \dfrac{1}{10}} = \frac{10|\alpha|}{9}(1 - 10^{n_0 - n_1}).$$
For any $n_0$, the magnitude of the roundoff error $\alpha$ at any step is certainly bounded by $y_{n_1 - 1} = \frac{1}{10n_1}$, so this can be substituted for $\alpha$ in the above equation.  The problem then is to show that $\exists n_1$ for any $\epsilon \in \mathbb{R}, n_0 \in \mathbb{N}, n_0 < n_1$ such that $|\frac{1}{9n_1}(1 - 10^{n_0 - n_1})| < \epsilon$.
A: The recursion may be numerically unstable because of repeated multiplications by $10$. However, if we delay the numerical part until we've used enough exact math, we get a nice series for $y_n$ that converges quickly.
If we multiply both sides of the recursion by $(-10)^{-n}$, we get
$$
(-10)^{-n}y_n-(-10)^{-n+1}y_{n-1}=\frac1{n(-10)^n}\tag{1}
$$
Since $y_0=\log(11/10)$, we get
$$
\begin{align}
(-10)^{-n}y_n
&=\log(11/10)+\sum_{k=1}^n\frac1{k(-10)^k}\\
&=-\sum_{k=n+1}^\infty\frac1{k(-10)^k}\tag{2}
\end{align}
$$
because $\sum\limits_{k=1}^\infty\frac1{k(-10)^k}=-\log(11/10)$.
Multiplying by $10^n$ and reindexing yields
$$
\begin{align}
y_n
&=-\sum_{k=n+1}^\infty\frac{(-10)^{n-k}}{k}\\
&=-\sum_{k=1}^\infty\frac{(-10)^{-k}}{n+k}\tag{3}
\end{align}
$$
and $(3)$ converges pretty rapidly (a bit over a digit a term).

Another Approach:
$$
\begin{align}
\int_0^1\frac{x^n}{x+10}\,\mathrm{d}x
&=\frac1{10}\int_0^1\left(x^n-\frac{x^{n+1}}{10}+\frac{x^{n+2}}{10^2}-\frac{x^{n+3}}{10^3}+\dots\right)\,\mathrm{d}x\\
&=\frac1{10}\sum_{k=0}^\infty\frac{(-1)^k}{(n+k+1)10^k}\tag{4}
\end{align}
$$
which is the same series as $(3)$ after reindexing.

Starting from your recursion
$$
\begin{align}
y_n
&=\frac1{10}\left(\frac1{n+1}-y_{n+1}\right)\\
&=\frac1{10}\left(\frac1{n+1}-\frac1{10}\left(\frac1{n+2}-y_{n+2}\right)\right)\\
&=\frac1{10}\left(\frac1{n+1}-\frac1{10}\left(\frac1{n+2}-\frac1{10}\left(\frac1{n+3}-y_{n+3}\right)\right)\right)\\[9pt]
&\dots\\[9pt]
&=\frac1{10}\frac1{n+1}-\frac1{10^2}\frac1{n+2}+\frac1{10^3}\frac1{n+3}-\dots+\frac{(-1)^{k-1}}{10^k}\left(\frac1{n+k}-y_{n+k}\right)
\end{align}
$$
which again leads to the same series once we notice that $y_n\le\frac1{10(n+1)}$ for all $n$.
A: $\displaystyle{y_{n} =\int_{0}^{1}{x^{n} \over x + 10}\,{\rm d}x:\quad ?}$
With $z \in {\mathbb C},\quad \left\vert z\right\vert < 1/10$:
\begin{align}
\sum_{n = 0}^{\infty}z^{n}y_{n}
&=
\int_{0}^{1}{1 \over 1 - zx}\,{1 \over x + 10}\,{\rm d}x
=
-\,{1 \over z}\int_{0}^{1}{1 \over x - 1/z}\,{1 \over x + 10}\,{\rm d}x
\\[3mm]&=
-\,{1 \over z}\,{1 \over 10 + 1/z}
\int_{0}^{1}\left({1 \over x - 1/z} - {1 \over x + 10}\right){\rm d}x
=
-\,{1 \over z + 10}
\left.\ln\left(x - 1/z \over x + 10\right)\right\vert_{0}^{1}
\\[3mm]&=
-\,{1 \over 10z + 1}
\left[\ln\left(1 - 1/z \over 11\right) - \ln\left(-1/z \over 10\right)\right]
=
\color{#ff0000}{\large%
-\,{\ln\left(10/11\right) \over 10z + 1}
-
{\ln\left(1 - z\right) \over 10z + 1}}
\end{align}
Now, expands the right hand side in powers of $z$. The coefficients are
$\left\{y_{n}\right\}$. 
$\color{#0000ff}{\large\mbox{We can easily see that the "offending terms"
will be the}\ \underline{\rm powers\ of\ \ 10}\,.}$
