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Question 0502b
Sigma Notation
Convergence II
Question
Generate new
A series
∑
r
=
1
n
u
r
{\displaystyle \sum_{r=1}^n u_r}
r
=
1
∑
n
u
r
is given by
∑
r
=
1
n
u
r
=
6
−
1
n
2
+
5
.
\sum_{r=1}^n u_r = 6 - \frac{1}{n^2 + 5}.
r
=
1
∑
n
u
r
=
6
−
n
2
+
5
1
.
What is the value of
∑
r
=
1
∞
u
r
{\displaystyle \sum_{r=1}^\infty u_r}
r
=
1
∑
∞
u
r
?
Attempt
∑
r
=
1
∞
u
r
=
{\displaystyle \sum_{r=1}^\infty u_r = }
r
=
1
∑
∞
u
r
=
Answer
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