Answer: (i) \(u_n=\frac{1}{\cos n}-\frac{1}{\cos(n-1)}\).
(ii) \(\sum_{n=1}^N u_n=\sec N-\sec 0=\sec N-1\).
(iii) The infinite series does not converge because the partial sums are \(S_N=\sec N-1\), and this does not tend to a limit as \(N\to\infty\).
(i) Start with the denominator and use the identity \(\cos P+\cos Q=2\cos\frac{P+Q}{2}\cos\frac{P-Q}{2}\).
Here \(P=2n-1\) and \(Q=1\), so
\(\cos(2n-1)+\cos 1=2\cos n\cos(n-1).\)
Now use \(\sin P\sin Q=\frac12\big(\cos(P-Q)-\cos(P+Q)\big)\) with \(P=n-\frac12\) and \(Q=\frac12\):
\(2\sin\left(n-\frac12\right)\sin\frac12=\cos\left(n-1\right)-\cos n.\)
So
\(4\sin\left(n-\frac12\right)\sin\frac12=2\big(\cos(n-1)-\cos n\big).\)
Substituting into the definition of \(u_n\),
\(u_n=\frac{2(\cos(n-1)-\cos n)}{2\cos n\cos(n-1)}=\frac{\cos(n-1)-\cos n}{\cos n\cos(n-1)}.\)
Split the fraction:
\(u_n=\frac{1}{\cos n}-\frac{1}{\cos(n-1)}.\)
(ii) Let \(S_N=\sum_{n=1}^N u_n\). Then
\(S_N=\sum_{n=1}^N \left(\frac{1}{\cos n}-\frac{1}{\cos(n-1)}\right).\)
Writing out the first few terms shows the cancellation:
\(S_N=(\sec 1-\sec 0)+(\sec 2-\sec 1)+\cdots+(\sec N-\sec(N-1)).\)
All the intermediate terms cancel, leaving
\(S_N=\sec N-\sec 0=\sec N-1.\)
Therefore
\(\sum_{n=1}^N u_n=\sec N-1.\)
(iii) For the infinite series to converge, the partial sums \(S_N\) must tend to a finite limit as \(N\to\infty\). But here
\(S_N=\sec N-1,\)
and \(\sec N\) does not approach a single value as \(N\to\infty\). In fact, it keeps changing and is not even bounded near values where \(\cos N\) is close to zero. Hence the partial sums do not converge, so the series \(u_1+u_2+u_3+\cdots\) does not converge.
