If for sufficiently large n, |(a_(n+1))/(a_n) |<k<1

If for sufficiently large $n$, $\left|\dfrac{a_{n+1}}{a_n}\right|

Explanation of Method

Once the ratio of consecutive terms is uniformly bounded by $k<1$, every later term is bounded by a geometric sequence starting at $|a_n|$; summing that geometric bound gives the remainder estimate directly.

Bound $|a_{n+j}|$ by $k^j|a_n|$ using the given ratio condition repeatedly.
Apply the triangle inequality to the remainder and sum the geometric bound.
Evaluate the resulting geometric series.

Step 1 · Bound successive terms

The remainder after $n$ terms is $R_n=\sum_{j=1}^{\infty}a_{n+j}$. From $|a_{m+1}|

Step 2 · Bound the remainder

By the triangle inequality, $$ |R_n| \le \sum_{j=1}^{\infty}|a_{n+j}| < \sum_{j=1}^{\infty}k^j|a_n| = |a_n|\sum_{j=1}^{\infty}k^j. $$

Step 3 · Sum the geometric series

Since $0

Summary Table

Property	Value
Hypothesis	$\|a_{n+1}/a_n\|
Term bound	$\|a_{n+j}\|
Result	$\|R_n\|<\|a_n\|\,k/(1-k)$

Additional Notes

This is the standard companion bound to the ratio test: whenever the ratio test proves convergence, this estimate immediately tells you how many terms are needed for a target accuracy, since $|R_n|=O(|a_n|)$.
For $\sum x^n/n!$ with $x=1$, the true ratio $1/(n+1)$ shrinks with $n$, so any fixed $k$ (say $k=0.01$) eventually bounds it once $n+1>1/k$; this makes the tail estimate extremely strong for factorially decaying series.
This bound underlies the practical error estimates used throughout Problems 279–280 and the numerical computations of Problems 283–288, where truncating a series after $N$ terms is justified precisely by an argument of this type.

Result

$$ \text{If } \left|\frac{a_{n+1}}{a_n}\right|