Question 1 · Sequences, series and the boundary
For \(n\ge1\), define
\[
a_n=\sqrt{n+4}-\sqrt{n+1}.
\]
(a) Show that \(a_n\to0\), and prove from the \(\varepsilon\)-\(N\)
definition that \(\sqrt n\,a_n\to3/2\). **(5 marks)**
(b) Determine whether \(\sum_{n=1}^{\infty}a_n\) converges. Your
justification must use partial sums rather than only the term test.
**(3 marks)**
(c) Prove that \(\sum_{n=1}^{\infty}(-1)^{n-1}a_n\) converges and
obtain an explicit condition on \(N\) that guarantees the truncation
error is below \(10^{-3}\). **(5 marks)**
(d) Let \(F(z)=\sum_{n=1}^{\infty}a_nz^n\). Find its radius of
convergence, then investigate separately \(z=1\), \(z=-1\), and
\(z=e^{i\theta}\) for \(0<\theta<2\pi\). **(7 marks)**
**[Total: 20 marks]**
Worked solution and marking guidance
**(a)** Rationalising gives
\[
a_n=\frac{3}{\sqrt{n+4}+\sqrt{n+1}}>0.
\]
In particular \(0<a_n\le 3/(2\sqrt{n+1})\), so \(a_n\to0\) (for
example, \(n>9/(4\varepsilon^2)\) is sufficient).
Also
\[
\sqrt n\,a_n=
\frac{3}{\sqrt{1+4/n}+\sqrt{1+1/n}}.
\]
Since \(\sqrt{1+t}-1\le t/2\) for \(t\ge0\),
\[
0\le\frac32-\sqrt n\,a_n
\le\frac34\left(\sqrt{1+4/n}-1+\sqrt{1+1/n}-1\right)
\le\frac{15}{8n}.
\]
Thus any integer \(N>15/(8\varepsilon)\) proves from the definition
that \(|\sqrt n\,a_n-3/2|<\varepsilon\) for all \(n\ge N\).
**(b)** The partial sums cancel in three shifted strings:
\[
\sum_{n=1}^N a_n
=\sqrt{N+2}+\sqrt{N+3}+\sqrt{N+4}
-(\sqrt2+\sqrt3+2).
\]
They tend to \(+\infty\). Hence the positive series diverges.
**(c)** The numbers \(a_n\) are positive, decrease because their
rationalised denominators increase, and tend to zero. The alternating
series therefore converges and
\[
|S-S_N|\le a_{N+1}
=\frac3{\sqrt{N+5}+\sqrt{N+2}}
\le\frac3{2\sqrt{N+2}}.
\]
Consequently \(N\ge2{,}249{,}999\) is a simple sufficient integer
condition for the error to be strictly below \(10^{-3}\).
**(d)** Since \(a_n\sim3/(2\sqrt n)\), the root test gives \(R=1\).
At \(z=1\) the series is the divergent positive series from (b). At
\(z=-1\) it is conditionally convergent by (c). For
\(z=e^{i\theta}\), \(0<\theta<2\pi\), the partial sums of
\(e^{in\theta}\) are bounded and \(a_n\downarrow0\), so Dirichlet's
test gives convergence. It is never absolute on \(|z|=1\), because
\(\sum |a_nz^n|=\sum a_n\) diverges.