An infinite tail from a supplied sum

Given ∑r=1nr+22r+1r(r+1)=12−1(n+1)2n+1\sum_{r=1}^n\frac{r+2}{2^{r+1}r(r+1)}=\frac12-\frac1{(n+1)2^{n+1}}, evaluate ∑r=5∞r+22r+1r(r+1)\sum_{r=5}^{\infty}\frac{r+2}{2^{r+1}r(r+1)}, leaving the answer as p/qp/q where p,qp,q are real. marks

[modified 2008/TPJC/Prelim/I/7]

Solution

The whole infinite sum

Limit of the partial sum

Given ∑r=1nr+22r+1r(r+1)=12−1(n+1)2n+1\sum_{r=1}^n\frac{r+2}{2^{r+1}r(r+1)}=\frac12-\frac1{(n+1)2^{n+1}}, find the sum to infinity ∑r=1∞r+22r+1r(r+1)\sum_{r=1}^\infty\frac{r+2}{2^{r+1}r(r+1)}.

Hint

  1. Let n→∞n\to\infty in the given formula.

Answer · The whole infinite sum

∑r=1∞r+22r+1r(r+1)=12\sum_{r=1}^\infty\dfrac{r+2}{2^{r+1}r(r+1)}=\dfrac12.

The tail from r = 5

Tail as a difference

Evaluate ∑r=5∞r+22r+1r(r+1)\sum_{r=5}^\infty\frac{r+2}{2^{r+1}r(r+1)}, leaving the answer as pq\frac pq.

Explore this idea

Hint

  1. Write the tail as the whole sum minus the sum of the first four terms.

  2. Use the given formula at n=4n=4.

Answer · The tail from r = 5

1/1601/160.

Explore the idea

The tail is tiny

The bars are the terms. Select the block from r=5r=5 onward and compare it with the formula Supper−Slower−1S_{\text{upper}}-S_{\text{lower}-1}.

Limits of the sumr = 5 to 14
Try an example
Σ (r+2)/(2^(r+1) r(r+1)), r = 5 to 14
0.006
Number of terms
10
S(upper) − S(lower − 1)
0.006
matches the direct total
Key idea

Almost all of the infinite sum, 12\frac12, is contained in the first four terms. What remains is the tail, 1160\frac1{160}.

Ideas to try
  1. Select the first four terms: the total is just below 0.50.5.
  2. Select r=5r=5 to 1414: the total is already almost exactly 1160=0.00625\frac1{160}=0.00625.
Takeaway

Whole minus head

When a result is only given for sums starting at 11, a tail is the infinite sum minus the head you need to skip.