(a) Consider \(w_{r+1} = (r+1)(r+2)\ldots(r+10)\) and \(w_r = r(r+1)(r+2)\ldots(r+9)\).

Then, \(w_{r+1} - w_r = (r+1)(r+2)\ldots(r+10) - r(r+1)(r+2)\ldots(r+9)\).

Factor out \((r+1)(r+2)\ldots(r+9)\):

\(= (r+1)(r+2)\ldots(r+9)((r+10) - r) = 10(r+1)(r+2)\ldots(r+9)\).

(b) The sum \(\sum_{r=1}^{n} u_r = \frac{1}{10}((w_2 - w_1) + (w_3 - w_2) + \ldots + (w_{n+1} - w_n))\).

This telescopes to \(\frac{1}{10}(w_{n+1} - w_1)\).

\(w_{n+1} = (n+1)(n+2)\ldots(n+10)\) and \(w_1 = 1 \cdot 2 \cdot \ldots \cdot 10 = 10!\).

Thus, \(\sum_{r=1}^{n} u_r = \frac{1}{10}((n+1)(n+2)\ldots(n+10) - 10!)\).

(c) The series \(v_1 + v_2 + \ldots + v_n = x^{w_2} - x^{w_1} + x^{w_3} - x^{w_2} + \ldots + x^{w_{n+1}} - x^{w_n}\).

This telescopes to \(x^{w_{n+1}} - x^{w_1}\).

For convergence as \(n \to \infty\), \(|x| < 1\) is required.

Thus, for \(-1 < x < 1\), the sum to infinity is \(-x^{w_1} = -x^{10!}\).

For \(x = \pm 1\), the sum to infinity is 0.

(a) The series \(S_n = \sum_{r=1}^{n} (x^{f(r)} - x^{f(r+1)})\) telescopes to \(x^{f(1)} - x^{f(n+1)}\).

(b) With \(f(r) = \ln r\), the series becomes \(\sum_{r=1}^{\infty} (x^{\ln r} - x^{\ln(r+1)})\). For convergence, \(0 < x \leq 1\). If \(x < 1\), the series converges to \(1\). If \(x = 1\), the series converges to \(0\).

(c) With \(f(r) = 2 \log_x r\), \(x^{2 \log_x r} = r^2\). Thus, \(S_n = 1 - (n+1)^2 = -n^2 - 2n\). Using standard results, \(\sum_{n=1}^{N} (-n^2 - 2n) = -\frac{1}{6} N(N+1)(2N+7)\).

(a) Consider \(\sum_{r=1}^{n} e^{rx}(e^{2x} - 2e^x + 1) = e^x - e^{2x} - e^{(n+1)x} + e^{(n+2)x}\).

Using the method of differences, the terms cancel out in a pattern, leading to the final result.

(b) For convergence, \(e^{nx} \to 0\) as \(n \to \infty\), which occurs when \(x < 0\). The sum to infinity is \(e^x - e^{2x}\).

(c) Using the laws of logarithms, \(\sum_{r=1}^{n} \ln u_r = \sum_{r=1}^{n} (rx + \ln(e^x - 1)^2)\).

Applying the formula \(\sum_{r=1}^{n} r = \frac{1}{2} n(n+1)\), we get \(\frac{1}{2} xn(n+1) + n \ln(e^x - 1)^2\).

(a) Start with the identity for the difference of tangents:

\(\tan(r+1) - \tan r = \frac{\sin(r+1)\cos r - \cos(r+1)\sin r}{\cos(r+1)\cos r}\)

This simplifies to:

\(\frac{\sin 1}{\cos(r+1)\cos r}\)

Thus, \(\tan(r+1) - \tan r = \frac{\sin 1}{\cos(r+1)\cos r}\).

(b) Using the method of differences, we have:

\(\sum_{r=1}^{n} u_r = \sum_{r=1}^{n} \frac{1}{\cos(r+1)\cos r}\)

Recognizing the telescoping nature:

\(\sum_{r=1}^{n} \left( \tan(r+1) - \tan r \right) = \tan(n+1) - \tan 1\)

Thus, \(\sum_{r=1}^{n} u_r = \frac{\tan(n+1) - \tan 1}{\sin 1}\).

(c) As \(n \to \infty\), \(\tan(n+1)\) oscillates, so the series \(u_1 + u_2 + u_3 + \ldots\) does not converge.

(a) Consider the sum \(\sum_{r=1}^{n} \left( \ln r - 2 \ln (r+1) + \ln (r+2) \right)\).

This expands to:

• \(\ln 1 - 2 \ln 2 + \ln 3\)
• \(\ln 2 - 2 \ln 3 + \ln 4\)
• \(\ln 3 - 2 \ln 4 + \ln 5\)
• \(\ldots\)
• \(\ln (n-1) - 2 \ln n + \ln (n+1)\)
• \(\ln n - 2 \ln (n+1) + \ln (n+2)\)

Most terms cancel, leaving:

\([\ln 1] - \ln 2 - \ln (n+1) + \ln (n+2) = \ln \frac{n+2}{2(n+1)}\).

(b) The sum to infinity is \(S = -\ln 2\).

We need \(S_n - S = \ln \left( \frac{n+2}{n+1} \right) < 0.01\).

This leads to \(n+2 < e^{0.01}(n+1)\).

Solving gives the least value of \(n\) as 99.

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.

Answer: (a) \(2n^4+8n^3+10n^2+7n\)

(b) \(1-\dfrac1{(n+1)^4}\)

(c) \(1\)

(a)

\(\sum_{r=1}^{n}(8r^3+12r^2+4r+3)=8\sum r^3+12\sum r^2+4\sum r+3n.\)

Using MF19,

\(8\left(\frac{n(n+1)}2\right)^2+12\left(\frac{n(n+1)(2n+1)}6\right)+4\left(\frac{n(n+1)}2\right)+3n\)

\(=2n^2(n+1)^2+2n(n+1)(2n+1)+2n(n+1)+3n.\)

Expanding and simplifying gives

\(2n^4+8n^3+10n^2+7n.\)

(b)

First put the left-hand side over a common denominator:

\(\frac1{r^4}-\frac1{(r+1)^4}=\frac{(r+1)^4-r^4}{r^4(r+1)^4}.\)

Since \((r+1)^4-r^4=4r^3+6r^2+4r+1\),

\(\frac1{r^4}-\frac1{(r+1)^4}=\frac{4r^3+6r^2+4r+1}{r^4(r+1)^4}.\)

Therefore

\(\sum_{r=1}^{n}\frac{4r^3+6r^2+4r+1}{r^4(r+1)^4}=\sum_{r=1}^{n}\left(\frac1{r^4}-\frac1{(r+1)^4}\right).\)

The terms telescope:

\(1-\frac1{2^4}+\frac1{2^4}-\frac1{3^4}+\cdots+\frac1{n^4}-\frac1{(n+1)^4}=1-\frac1{(n+1)^4}.\)

(c)

From part (b), the partial sum is \(1-\frac1{(n+1)^4}\). Hence

\(\lim_{n\to\infty}\left(1-\frac1{(n+1)^4}\right)=1.\)

Mark scheme

Answer: (a) \(\dfrac14 n(n+1)(n-1)(n+2)\)

(b) \(\dfrac32-\dfrac2n+\dfrac1{n+1}\)

(c) \(\dfrac32\)

(a)

\(\sum_{r=1}^{n}(r^3-r)=\sum_{r=1}^{n}r^3-\sum_{r=1}^{n}r.\)

Using MF19,

\(\sum r^3=\left(\frac{n(n+1)}2\right)^2,\qquad \sum r=\frac{n(n+1)}2.\)

Hence

\(\left(\frac{n(n+1)}2\right)^2-\frac{n(n+1)}2=\frac14n(n+1)\{n(n+1)-2\}.\)

Since \(n(n+1)-2=n^2+n-2=(n-1)(n+2)\), the answer is

\(\frac14 n(n+1)(n-1)(n+2).\)

(b)

Since \(r^3-r=r(r-1)(r+1)\), write

\(\frac{r+3}{r(r-1)(r+1)}=\frac{A}{r-1}+\frac{B}{r}+\frac{C}{r+1}.\)

Multiplying by \(r(r-1)(r+1)\) gives

\(r+3=Ar(r+1)+B(r-1)(r+1)+Cr(r-1).\)

Solving gives \(A=2, B=-3, C=1\), so

\(\frac{r+3}{r^3-r}=\frac2{r-1}-\frac3r+\frac1{r+1}.\)

Therefore

\(\sum_{r=2}^{n}\left(\frac2{r-1}-\frac3r+\frac1{r+1}\right)\)

has cancelling terms, leaving

\(\frac32-\frac2n+\frac1{n+1}.\)

(c)

From part (b),

\(\sum_{r=2}^{n}\frac{r+3}{r^3-r}=\frac32-\frac2n+\frac1{n+1}.\)

Letting \(n\to\infty\) gives

\(\frac32.\)

Mark scheme