Problem: The limiting sum of the infinite series, 101​+1022​+1033​+… whose nth term is 10nn​ is:
Answer Choices:
A. 91​
B. 8110​
C. 81​
D. 7217​
E. larger than any finite quantity
Solution:
Write the given series as:
101​+1021​+1031​+1041​++1021​+1031​+1041​++1031​+1041​+​……… and so forth​
Let s1​ be the limiting sum of the first-row series, s2​, that of the second-row series, and so forth.
s1​s2​s3​​=1−101​101​​=91​,=1−101​1021​​=901​,=1−101​1031​​=9001​,and so forth.​
and so forth.
Therefore, the required limiting sum equals
91​+901​+9001​+…=1−101​91​​=8110​