4.4.3 Sum of the Reciprocals of Triangular Numbers

This formula is independent of knowing the triangular numbers themselves – you just need to know which term number they are.

The Formula

1Tn+1Tn+1+⋯+1Tm=2(1n−1m+1)\frac{1}{T_n} + \frac{1}{T_{n+1}} + \dots + \frac{1}{T_m} = 2 \left( \frac{1}{n} - \frac{1}{m + 1} \right)

Example: Starting from T₁

1+13+16+110=2(11−15)=2×45=851 + \frac{1}{3} + \frac{1}{6} + \frac{1}{10} = 2 \left( \frac{1}{1} - \frac{1}{5} \right) = 2 \times \frac{4}{5} = \frac{8}{5}

Example: Starting from T₃

16+110+115+121=2(13−17)=821\frac{1}{6} + \frac{1}{10} + \frac{1}{15} + \frac{1}{21} = 2 \left( \frac{1}{3} - \frac{1}{7} \right) = \frac{8}{21}

Triangular Numbers Reference

nT_n
11
23
36
410
515
621
728

Problem Set 4.4.3

Practice these problems. Type your answer and press Enter to check:

1/3 + 1/6 + 1/10 + 1/15
1/3 + 1/6 + 1/10
1/3 + 1/6 + 1/10 + 1/15 + 1/21
1/6 + 1/10 + 1/15 + 1/21
1/6 + 1/10 + 1/15