∑n=1infinity4/(n+3)\sum_{n=1}^{infinity}{4/(n+3)}

asked by guest
on Nov 24, 2024 at 8:37 am



You asked:

Evaluate the expression: ∑n=1∞4n+3\sum_{n = 1}^{\infty} {\frac{4}{n + 3}}

MathBot Answer:

The infinite series ∑n=1∞4n+3\displaystyle\sum_{n=1}^{\infty} \frac{4}{n + 3} diverges.

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